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Commercial drain cleaners sometimes contain mixture of aluminum powder and NaOH: When the mix is poured down clogged drain, lhe following occurs: NaOH(aq) 2AI(s) 6H...

Question

Commercial drain cleaners sometimes contain mixture of aluminum powder and NaOH: When the mix is poured down clogged drain, lhe following occurs: NaOH(aq) 2AI(s) 6HRO() 2NaAI(OHA(aq) 3Hi(g) HeatHeat melts grease and gas breaks up the clogs_ What volume of hydrogen formed at 00 atm and 21.09C when of Al are treated with excess NaOH?- t '1 ~aS3'Opih"<-The mass percent of HCOs in a certain 3.20 g Alka Seltzer tablet is 32.5%. How many mL of carbon dioxide will be generated when t

Commercial drain cleaners sometimes contain mixture of aluminum powder and NaOH: When the mix is poured down clogged drain, lhe following occurs: NaOH(aq) 2AI(s) 6HRO() 2NaAI(OHA(aq) 3Hi(g) Heat Heat melts grease and gas breaks up the clogs_ What volume of hydrogen formed at 00 atm and 21.09C when of Al are treated with excess NaOH? - t '1 ~a S 3 'Opih" <- The mass percent of HCOs in a certain 3.20 g Alka Seltzer tablet is 32.5%. How many mL of carbon dioxide will be generated when the Alka Seltzer reacts with the HCI in person = stomach at a pressure 780 torr and temperature 37.0AC? HCOs HCI(aq) Cl(aq) HzO() COz(g) 034)Mu JA5 nana '7.f "1 ' ' What mass of NaNs is needed to inflate an 18.0 L air bag in an automobile at 25.09C and 1.00 atm according to the following reaction? GNaNa(s) FezOz(s) 3NazO(s) 2Fe(s) 9Nz(g) 1 [' ; J = TJnuteaiy 1 =4 ._n auerem 7Z 6 . #onf ~' '11'' (h _ F



Answers

Some commercial drain cleaners contain a mixture of sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: $$\begin{aligned}2 \mathrm{NaOH}(a q)+2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \\2\mathrm{NaAl}(\mathrm{OH})_{4}(a q) &+3 \mathrm{H}_{2}(g)\end{aligned}$$ The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of $\mathrm{H}_{2}$ formed at $23^{\circ} \mathrm{C}$ and 1.00 atm if $3.12 \mathrm{g}$ of $\mathrm{Al}$ are treated with an excess of NaOH.

Okay, this is question 108 from chapter 11. And we're trying to find the volume hydrogen when we have a 3.12 grams aluminum. So the first thing we do is find the most of package in. So they said we have three might want to your crafts of a woman, and we're gonna multiply that by the most. From all the massive aluminum shell. One more of aluminum is equal 2 27 grams as a limited. Um, and then we're gonna move. Multiply that by the molar ratio between, um, hydrogen, an aluminum. So we have a lot of the equation. We have three most of hydrogen compared to the two moves of aluminum. And this is going to come out to be 0.1 73 Most of hydrogen. Okay, now that we have that, we can find the value. And volume is dr found. That V is equal to energy. T overpay the ideal gas, ma. Okay, so we have 0.173 moles. Time is the gas constant. That's temperature and coven. They give us 23 degrees Celsius. So at 273 to come in 3 296 coverage over the pressure, which is Ah one. And this is gonna come out to be a volume of four 0.20 leaders.

Okay, so let's start by finding the polarity Of our Nuh, we know we have 4992 g. We'll go ahead and change that to malls smaller, massive NHS 40. So it gives us 1.248 moles. So then our polarity is moles per liter of solution. So we've got 1.248 moles And that's dissolved in .6 L of solution. So it's 2.08 moller nuh, So we're gonna figure out now how much hydrogen gas is being produced. But this is a limiting reactant problem. I know this because I'm giving up the amounts of both reactant. So let's go ahead and start with what we know about our Nuh, we found the malls up here. Let's go ahead and do a mall ratio. We'll convert moles of sodium hydroxide, two moles of hydrogen. And you can see from our equation, there's three moles of hydrogen for every two moles of O H minus. And that came from the Noh, So that's 1.872 moles of hydrogen. But we'll have to see what are limiting reactant is. I have 41-8 g aluminium was added to this. So we'll change grams to moles and then malls to malls, so grams of aluminum, two moles of aluminum, and then we'll change molds of aluminum, two moles of hydrogen. This ratio is also 3-2 And our molar mass of aluminum is 26.98. So this is going to give us to .30 moles of hydrogen. So we have enough aluminum to make 230 moles of hydrogen. But we only have enough sodium hydroxide To make this much of the hydrogen. sodium hydroxide is are limiting reactant. Are aluminum was our excess and our answer here is 1.872 moles. Okay, so now we're going to try to figure out how many leaders of gas that would be. So we're gonna use our ideal gas law PV equals NRT or B equals NRT overpay. So we've got our moles of hydrogen, the 1872 Ours point away to one. Our temperature was 25°C, which is 2 98 Kelvin. And then we need to divide it by our pressure. Well, Our total pressure was 758 0.6. But we need to subtract the vapor pressure of the water and that will give us the pressure of the hydrogen in millimeters of mercury. And then we'll go ahead and change that to ATM So 1/7 60. So our pressure is .96780 M. So we'll go ahead and put that that up here, .967 And we'll get that are volume is going to be 47 0.4 millimeters.

Let us calculate the volume of hydrogen gas from the given equation which is the reaction of aluminum with sodium hydroxide. The balance the equation is to a L. Place to a nearby village place six H to war. Use rice too. To any l. Watch four times placed three hits 2 from this as atomic mass of aluminum is 27 2 into 27 g of aluminum liberate sir. Three into two g of hydrogen. Yeah. Uh huh. Let us calculate the weight off my rosen liberated from 6.5 g of aluminum. Mhm. Yeah. Yeah. Way to fight. Rosen is equal to 6.5 into three into 2 x two into 27. Yeah. Mhm. Yeah. Okay. All right. Mhm. The value equals to 0.722 g. Yeah Moniker weight of fighters in these two g per mole. Yeah. Yeah. Yeah. Yeah. Yeah. By using the weight of hydrogen and molecular weight of hydrogen. We can calculate the number of moles number of moles of hydrogen is equal to the weight of high treason. By molecular weight of high treason Weight is 0.722 divide by molecular weight is too equals two 0.361 moles. Well right. The given data as the pressure is 7 42. Which can be converted into atmospheres 7 42 by 7 60 equals to 0.9763 atmospheres. Yeah. Yeah. Yeah. Yeah. The temperature is given as a 22 degrees Celsius which can be converted into Calvin 22 plus 2 73.15 equals 2 to 95.15 Calvin. Yeah. Yeah. Yeah. Yeah. The volume of fighters and gas liberated there can be calculated by using the formula P. V. Is equal to an R. P. Yeah. Yeah. Yeah. Yeah. From this we is equal to an arty by P. Yeah. Yeah. Sub shooting all the values number of moles is 0.361 malls into the value of gas constant. R. S. 0.0 81 leaders. Atmospheres killed in inverse. Okay. Mhm. Yeah. Yeah. Into temperature is 2 95.15 Calvin. Yeah. Yeah divide by The pressure is 0.9763 at most meals. Okay. Atmospheres atmospheres council moral universe mole. Council killing universe kelvin council the value equals two 8.96 leaders. Yeah. Which is close to nine leaders. So the volume of fighters and gas produced during the reaction is nine L.

Okay if you've got to aluminum combining with to any O. H. And six waters you're gonna form too. And a A. L. O. H four and then to hydrogen gas is so here we've got 6.5 g of aluminum. The sodium hydroxide is an access and then we've got To solve for the volume of hydrogen gas. And leaders they gave us a pressure of 742 mm of mercury and then they gave us the temperature of 22°C, which is 295 Kelvin. So start with your g, convert your g over two moles by dividing by 26.98, which is the molar mass of aluminum. The mole ratio over to the hydrogen. Using the coefficients You're gonna get .24 moles of hydrogen then use PV equals NRT to solve for volume. So we're gonna use this information. This is my N. And then you're going to use this information here. So this is 0.24 moles times the ideal gas constant 0.8 to one times 2 95. Divided by you've got to convert this millimeters of mercury to a. T. M. By dividing by 7 60. So this is actually, I'm just gonna put that in here like this. Okay, so here's your kelvin, here is your moles and here is your leader times atmosphere over more times kelvin. So kelvin will cancel um your A. T. M will cancel. Your moles will cancel leaving you with leaders. So we have six leaders


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