Question

1.) CH4(g) + Cl2(g) CHCl3(l) + HCl(g) What mass of HCl can be produced from 2.60 g of methane and 10.0 g of chlorine in the above unbalanced reaction? Use at least as many significant figures in your molar masses as in the data given. 2.) 21.0 g of aluminum and 231 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equation for the reaction is the following. 2 Al(s) + 3 Cl2(g) ? 2 AlCl3(s) What is the quantity of chlorine gas left, in grams, after the reaction has occurred, assuming that the reaction goes to completion? (The formula mass of aluminum metal, Al, is 26.98 g/mol, and the formula mass of chlorine gas, Cl2, is 70.90 g/mol.)

          1.)
CH4(g) +
Cl2(g)  CHCl3(l) +
HCl(g)
What mass of HCl can be produced from 2.60 g of methane and 10.0
g of chlorine in the above unbalanced reaction? Use at
least as many significant figures in your molar masses as in the
data given.
 
2.)
21.0 g of aluminum and 231 g of chlorine gas
react until all of the aluminum metal has been converted to
AlCl3. The balanced equation for the reaction is the
following.
2 Al(s) + 3 Cl2(g) ? 2
AlCl3(s)
What is the quantity of chlorine gas left, in grams, after the
reaction has occurred, assuming that the reaction goes to
completion? (The formula mass of aluminum metal, Al, is 26.98
g/mol, and the formula mass of chlorine gas, Cl2, is
70.90 g/mol.)

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Added by Brandi C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ashley Brady

04/20/2022

Step 1

) First, we need to balance the chemical equation: CH4(g) + 3Cl2(g) -> CHCl3(l) + 3HCl(g) Now, we can find the moles of each reactant: Moles of CH4 = (2.60 g) / (16.04 g/mol) = 0.1621 mol Moles of Cl2 = (10.0 g) / (70.90 g/mol) = 0.1410 mol Show more…

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1.) CH4(g) + Cl2(g) CHCl3(l) + HCl(g) What mass of HCl can be produced from 2.60 g of methane and 10.0 g of chlorine in the above unbalanced reaction? Use at least as many significant figures in your molar masses as in the data given. 2.) 21.0 g of aluminum and 231 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equation for the reaction is the following. 2 Al(s) + 3 Cl2(g) ? 2 AlCl3(s) What is the quantity of chlorine gas left, in grams, after the reaction has occurred, assuming that the reaction goes to completion? (The formula mass of aluminum metal, Al, is 26.98 g/mol, and the formula mass of chlorine gas, Cl2, is 70.90 g/mol.)

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Transcript

00:01 Okay, so you are given the following balanced chemical reaction.

00:04 You are also told that you are starting with 22 grams of aluminum, and you are starting with 27 grams of chlorine gas.

00:17 So part a says you have excess chlorine gas.

00:21 So it's wondering how many moles of al2 -c -o -3 you could produce.

00:27 Or al -c -l -c -l -3, sorry, you could produce.

00:31 So if we're going to go from grams to moles, that's going to be a two -step.

00:38 So if we do that, we go 22 grams of aluminum.

00:44 First thing to do is get rid of those grams of aluminum.

00:46 So grams of aluminum will go on the bottom.

00:48 The only thing that can go with grams of aluminum is one mole of that substance.

00:53 We're going to find the grams on the periodic table.

00:56 The mass for aluminum is 26 .98.

01:01 The second step then is to get rid of our moles of aluminum and get to our moles of alcl3.

01:10 When we have mole on top of mole, you have to look at our coefficients.

01:12 So we have a 2 for alcl3, and we have a 2 for al.

01:19 Since those two are the same, we can either simplify it to be 1 in 1, or we can just keep it as 2 and 2.

01:24 When i do my calculation, though, i don't really need to include the 2s because 2 divided by 2 is just 1...

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