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Calculate how many grams of oxygen form when 22.4 mg of KCIO3 reacts completely given the following equation:KCIO3 () 2 KCI (s) 3 02 (0)...

Question

Calculate how many grams of oxygen form when 22.4 mg of KCIO3 reacts completely given the following equation:KCIO3 () 2 KCI (s) 3 02 (0)

Calculate how many grams of oxygen form when 22.4 mg of KCIO3 reacts completely given the following equation: KCIO3 () 2 KCI (s) 3 02 (0)



Answers

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts. $$2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ (a) $2.72 \mathrm{g} \mathrm{KClO}_{3}$ (b) $0.361 \mathrm{g} \mathrm{KClO}_{3}$ (c) $83.6 \mathrm{kg} \mathrm{KClO}_{3}$ (d) $22.4 \mathrm{mg} \mathrm{KClO}_{3}$

In this question. We have magnesium and oxygen gas reacting on and were asked how many grams of magnesium oxide will form when 14.8 liters of oxygen gas measured on STP reacts with my museum metal. So STP is standard temperature and pressure and that means the temperature of 2 73 Calvin and a pressure of one atmosphere. So then the first thing we're gonna do is find out how many moles of oxygen are in this 14.8 liters of gas. So let's do that. I'm gonna use the ideal gas equation which is PV equals and r t where p is the pressure which we have here. The is the volume which we have here and is what we're looking for. That's the number of moles are is a constant and t is temperature which we have right here. So we plug all these numbers and we have one atmospheres or pressure 14.8 leaders as our volume and is that we're looking for are is a constant. It's equal 2.0, a 21 leader atmosphere her mole Kelvin And then the temperature is to 73. OK, so we do the math here and solve for end. We will get an end equal. Teoh 0.66 Moles of oxygen. So that's how many moles of oxygen we had in that 14.8 leaders of guests. Now, if I want a good, um, grams of magnesium oxide, I need to first go two moles of the museum oxide. So for every one mole of 02 I have two moles of my museum oxide according to our balanced equation up here, Um, so moles of 02 will cancel out. And that will give me 1.32 moles of my museum oxide and then do 2 g. I will just take that and multiply it by the molar mass of museum oxide. So 1.32 bills, many my museum oxide times 40.3 g per mole. Moles cancels out and I get 53.2 g of Thank you, my friend.

In this question, we have zinc metal reacting with oxygen, gas forming zinc oxide. Um We end up with one g of zinc oxide and we need to calculate in order to end up with one g of zinc oxide. What is the mass in grams of oxygen that we started off with? Right, so this is a mass mass storage geometry type of problem. First of all, we need to convert the mass of zinc oxide to the number of moles of zinc oxide. And then we can calculate the number of models of oxygen by using the moderation. And from there, it's easy then to get the mass of oxygen that we needed To produce the one g zinc oxide. All right, so let's do this. The number of moles of zinc oxide is equal to the one grand that we started off with divided by the molar mass of zinc oxide and that is 81 81.39 g per mole right from you. We need to go um calculate the number of moles of oxygen by using the mole ratio. So moles of oxygen over moles off zinc oxide is equal to. Now, let's check the stock geometry coefficients here for oxygen. It's one for zinc, it's a zinc oxide. It's too, so it's 1/2, one over to. Therefore the number of moles of oxygen is equal to a half times the number of moles of zinc oxide, which we have calculated in the step above. And now from this step, we can say therefore the mass of oxygen is going to be that half times the number of moles of zinc oxide, times the molar mass of oxygen, and the molar mass of oxygen Is 32.00 g per mole. Right, So let's just substitute all the values here so that we can do the complete calculation. So a number of moles of zinc oxide, we see it's one g one g divided by, oops, sorry about that. Divided by the molar mass. I wasn't oxide, which is 81.39 g per mole. And then we need to multiply that by 32.00 g per mole, which is the molar mass of oxygen. If we do that calculation, we end up With a massive oxygen of 0.20 grams. Right? So just to go over this again, we have a mass mass type of geometry problem here, we've been given the mass of the product, so therefore we need to convert it to the number of moles of product first, which we can then use to calculate the number of moles of oxygen, which is one of the reactant. And we do this by using the mole ratio. And then once we have the number of moles of oxygen, we calculated the massive oxygen by multiplying the number of moles of oxygen by its molar mass.

Using the equation provided for eight. Calculate the volume of oxygen that can be made from five 0.0 times 10 to the minus. Two malls of K C L 03 usar strike geometry two moles of K C. L 03 23 Moles of oxygen. One mole of oxygen is equal to molar mass at 32 g of oxygen. Then the density here is 1.4 to 8 grounds per leader and this would yield 1.68 leaders of oxygen. So this would be the volume of oxygen that could be made from 5.0 times 10 to the minus. Two moles of K C l 03 for part B. How many grams of K. C. L. 03 must react to form 42.0 mL of oxygen. So starting from 42.0, the leaders of oxygen that dense Let's first convert this to leaders 1000 mill leaders per leader. The density of oxygen is 1.4 to 8 grams per leader, 32.0 g of 02 In one mole, three moles of 02 to two moles of cassio, three and then times one more K c 03 This would have a mass of 122.6 grams is the smaller mouse. This would work out to 0.153 grams of K c L 03 and for part C how many milliliters of oxygen will form an STP from 55 0.2 g of Casey Low three. So starting from 55 0.2 g of K. C. L 03 20.6 grounds in one more two moles of K C l 03 the three moles of 02 and at STP one more, level two, it's equivalent to 22.4 liters of oxygen. This would be equal to 15.1 Leaders of oxygen at STP

Yeah, yeah, they're given a reaction. Any cost Number 31 Do z o giving to HD prosciutto. Yeah. Know how many grams of old skin form when each quantity off the reaction completely reacts. So the first country given is, uh, two point Monte Grande 2.13 g. What? We all sail in second days. The 6.77 Um what gold side in third case 1.55 radio maybe 24 days. People in 87 millions. Now let us see. What we have to do is are we finally more a mosque off two months off Mercury culture Mercury cultured is having a monocle. Mama, I must have 216 point for your name. So to times 16.59 guns will reduce 32 grants. A positive. That's what depression. So did you land up off? So when 2.13 grand reacts now, the factor between these two this 32 aboard to induce 2.216 point 59 equals 0.74 So that is the fact that we multiply this factor with each of these. Number two bond 13 in 20 point 074 Will give mhm 0.157 0.15 7 g off oxygen They reproduce when toe full Monty Gambia's again here also be multiplied the same with 0.74 every day. 0.5 g graham oxygen on here also, we do the same thing we will get like if they bite zero point you get 0.11 Quite easy. Andi In the last one when we were deprived the same factor we get 0.286 mg. 286 mg which even writers you 0.86 independent okay Or minus war? Yes, possibly.


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