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How many grams of hydrogen gas are produced from the reaction of 5.4 grams of aluminum with 25 g of HCI? 2 Al(s) 6 HCI(aq) ~- > 2 AICls(aq) 3 Hz(g)0.30 g0.60 g0....

Question

How many grams of hydrogen gas are produced from the reaction of 5.4 grams of aluminum with 25 g of HCI? 2 Al(s) 6 HCI(aq) ~- > 2 AICls(aq) 3 Hz(g)0.30 g0.60 g0.69 g8.1gWhich one of the following atoms has the largest radius?

How many grams of hydrogen gas are produced from the reaction of 5.4 grams of aluminum with 25 g of HCI? 2 Al(s) 6 HCI(aq) ~- > 2 AICls(aq) 3 Hz(g) 0.30 g 0.60 g 0.69 g 8.1g Which one of the following atoms has the largest radius?



Answers

Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and $35.8 \mathrm{~mL}$ of hydrogen gas over water at $27^{\circ} \mathrm{C}$ and $751 \mathrm{mmHg} .$ How many grams of aluminum react?

Okay, so today we're going to look at the law of the conservation mass and the law of conservation of mass says that mass cannot be created or destroyed. It simply changes form. If you look at the bottom of this diagram or this image, you'll see it right here. It means it cannot be created nor destroyed. It can only change forms. Okay, so we're looking at a chemical equation and put an arrow right here, so we can see what's going on. And in this chemical equation we have hydrochloric acid and aluminum foil and it is producing a solution hydrogen gas, but there is aluminum foil leftover. So we kind of say it produces aluminum foil. Anyway, we're looking at the conservation of mass. So we're saying that the mass of the initial part of the equation is going to be equal to the mass you get in the final stage of the equation. And this applies and is more easily figured out. Of course, when you're looking at a closed system, why? Because if it's an open system, that means that that hydrogen gas that's being produced could go out into the air until you get two different masses. It doesn't mean that you've lost the mask. Just means that the mass has gone out in the air. But we're gonna assume this is a closed system and were able to keep everything that happens to be made. Okay? All right, So, we need the mass of our initial Reactant. Remember your reactions are always going to be on your left, products are always going to be on your right. So, if we look at our reactant, we find that we have 18 g of aluminum foil and we have wait a minute. Not grams. We have 25 mL of hydrochloric acid. Well, that's not going to work at all. We have to do something about that. So, we know we have 25 mL of hydrochloric acid. We need that to be changed into grams. So look at your formulas at the bottom. Ha right here it gives us the density of hydrochloric acid is 1.25 g per centimeter cube. Well that's not milliliters. No, you're forgetting. Make another set of Astra. I love Astra's another set of Astra's. Remember that centimeter cubed is equal to milliliters. So however minutes, centimeter cubes you have, you have the same number of middle leaders. So that's the same. So I'm just gonna go here just so I don't throw you off. And I'm going to turn this into middle leaders because that's what it actually is. So now I can do the conversion, Remember that. Any number is over one. And so if I set up a ratio to convert this, since I have middle leaders in my numerator, I'm gonna need to put my middle leaders in my denominator so I can cancel them out, right? And this is one millimeter And it is 1.025g. So if I cancel out my ml, which I can now And I take my trusty dusty calculator and I take 25 and multiply it by 1.025. I'm gonna get 25 .6-5 g of hydrochloric acid. Okay, That works for me and put a little thing around this because I just calculated that out. And then I'm gonna go over here and I'm gonna add that because you're adding that 25 point 6 to 5 g. That's going to tell me the mass of everything that's in my reactant, I add those two together and that's gonna give me 43 .625 ramps. Okay, so now I know what my initial mass is. Right, okay, then I look at my Products. Well, it tells me that I've got 12 g of aluminum foil. That's fine. And I've got 30 .95 g of solution, whatever that is. But I don't have my hydrogen gas. I gotta figure out how much it's like, well, that should be easy enough. I just need to find out how much I've made so far. So what I have and whatever is left will be hydrogen. Of course. So I'm gonna take my 12 g and added to my 30.95 g. And that's going to give me what, 40 two point 95g. Okay, so that gives me that and that just gives me the aluminum solution. So, to find out my amount of hydrogen, give yourself a little bit more working room, find out my amount of hydrogen. I'm going to have to take my Initial amount of 43 .625 and subtracted. So trapped 42 .95 from it. So when I do that subtraction, I'm going to wind up with 0.6 75 g Left. Now this was everything. And the 42.95 was the products that made so far. So this 0.67 five must be the hydrogen gas that's left. So hey, I've got my massive hydrogen gas. I should be able to put it back up here. But oh no, no, no, no, no. If I'm looking for leaders of hydrogen gas, well, I don't have leaders. I've got grams of hydrogen gas so I'm gonna have to do something. I fixed that. So let me look up. There we go right there. Density of hydrogen gas is 0.8 to four g per liter. And I need leaders. So that's right there. So I can use that as my conversion factor. So I'm going to take 0.675 g of hydrogen gas. Remember that's going to be over one times my conversion factor, which is programs on the bottom 0.8 to four g to everyone. Leader. That way, I can cancel out my grams. And on my calculator, 0.675 divided by 0.8 24 is going to give me eight 192 leaders. And I'm talking about H two and I wanted to know the leaders of H two. And so there is my answer. Okay, so recap very quickly what we did was we added up the mass of our reactant, We added up the mass we had of our products. Then we found the difference between the two to find out how much was hydrogen and hydrogen was given to us in grams, but we couldn't use grams. So we had to take that hydrogen. Now there's the grams of hydrogen recalculated and take that hydrogen, and we have to convert that into leaders using the density of hydrogen gas, which gave us the volume of hydrogen gas and leaders, which is what you wanted. And that ends your lesson.

From the data given. Let's calculate how many grams of aluminum reacted. Let's first start with our balanced equation. We're told that aluminum it's going to react with hydrochloric acid to produce aluminum chloride, Equus, aluminum chloride and hydrogen gas to balance his equation, uh, to aluminum six hcl to L. C. L three on three. H two told that the HCL is president in excess, 35.8 mL of Hadron Gas, um, over water at 27 degrees Celsius and 751 million m of mercury. So let's first solve for the moles of hydrogen. Let's first convert our 35.8 mL of each to into leaders 1000 little leaders in one leader. This would be 0.358 Leaders of H two 27 degrees Celsius is equivalent to 300. Calvin had to 73 to that, and 751 millimeters of mercury will convert this to atmospheres. Uh, there's 760 millimeters of mercury in one atmosphere, so this is equal to 0.988 atmospheres. So, using this, let's use our ideal gas, um, formula and calculate moles of thistles moles of hydrogen pressure is 0.988 atmospheres. The volume is 0.358 leaders Ideal gas constant 0.8 to 06 leaders atmospheres per mole. Kelvin temperature is 300 Calvin. So solving this, we find that we have 2366.144 moles of hydrogen. So now we can apply our story Geometry 0.144 moles of hydrogen From our balance equation we see that we have three moles of hydrogen to two moles of aluminum. One more of aluminum has a Moeller Massa 27 0 g and solving. Here we find that this is equal to 00259 g of aluminum. So this would be the mass of aluminum that did react.

Okay, so this equation is not balanced. You have got three chlorine and two hydrogen on the product side. And since hydrogen chlorine are together in hydrochloric acid on the reactant side, the lowest common multiple of three and two A six. So I would place a six in front of hcl. So since you have six hydrogen in six chlorine, now put it to here To balance the chlorine and a three here to balance the hydrogen. Now I've got to aluminum's, we're gonna put it to in front of the aluminum. So your coefficients would be 6-3. Now, you're originally given 0.466 g of aluminum and the concentration of hcl is 0.100 polarity. So in part a we are going to be solving for the leaders of hydrogen gas produced at STP. So we want to write up some sort of a flow chart to help us get there. So, if we have grams of aluminum, you can convert grams of aluminum, two moles of aluminum by dividing by the molar mass. And then you can convert moles of aluminum, two moles of hydrogen by using the coefficients in the balanced equation, doing a mole ratio. And then you can convert to leaders of hydrogen by using 22.4 cents were at STP. So if you use the .466, let me see if I can fix this here, Change the color for you. If we can use the .466g of aluminum and we're going to divide this by the molar mass of aluminum, which is 26.98 from the periodic table. We convert to moles. So these units will cancel. And then we can convert our moles of aluminum by using the two coefficient in front of the aluminum and the balanced chemical equation. And we can convert two moles of hydrogen. So this too is now over here and this three is right here. Then, since we're at STP, we can convert to Leaders using 22.4. Yeah, if you put all of this in your calculator and you wanna probably round this off to three significant figures because that's the fewest that you have. You're going to end up with point 580 leaders of hydrogen. Okay, moving on to part B. We are going to be solving for the leaders of hydrochloric acid. And in order to do this, if we write up a plan to get there, if we start with our grams of aluminum, you know, we can convert two moles of aluminum like we did in part by dividing by the molar mass. And then you can convert moles of aluminum, two moles of hydrochloric acid using the coefficients from the balanced equation. And then you can convert to leaders of hydrochloric acid by using the polarity. So let me change the color Here. And we're gonna use .466g of aluminum, Divide this by the Molar Mass 26 98 from the periodic table. These units will cancel. Then we're going to use the coefficient in front of aluminum in the balanced chemical equation and the coefficient in front of hydrochloric acid. These units cancel. And now we can use the polarity of hydrochloric acid. So since it's .100 Moles, I'm sorry polarity that's the same thing as .100 moles of hydrochloric acid per liter. So these units here will cancel, leaving us with leaders. And again, we're going to round this off to three significant figures since that's the fewest sig figs that we've got. And That's gonna give us .518 leaders.


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