Given a percent yield of 45.0% for the reaction:

H2 + Cl2 → 2 HCl(g)
How many moles of HCl gas are produced
if 15.5 L of Cl2 at STP and excess H2 are
reacted?
1. 0.769 mol
2. 0.346 mol
3. 0.622 mol
4. 3.08 mol
5. 1.38 mol
6. 0.156 mol

0.622 mol

To answer this question, we need to follow a step-by-step calculation using the given information and the concept of percent yield.

Step 1: Convert the given volume of Cl2 to moles.
We are given the volume of 15.5 L of Cl2 at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies 22.4 L. Therefore, we can calculate the number of moles of Cl2 using the following formula:

moles of Cl2 = volume of Cl2 / molar volume at STP
moles of Cl2 = 15.5 L / 22.4 L/mole
moles of Cl2 = 0.693 moles (rounded to three decimal places)

Step 2: Use the stoichiometry of the balanced equation to find the moles of HCl produced.
From the balanced equation: 1 mole of Cl2 reacts to produce 2 moles of HCl.
Therefore, if we have 0.693 moles of Cl2, we will produce twice as many moles of HCl. Therefore:

moles of HCl = 2 * moles of Cl2
moles of HCl = 2 * 0.693
moles of HCl = 1.386 moles (rounded to three decimal places)

Step 3: Calculate the actual yield using the percent yield.
The percent yield given is 45.0%. Percent yield is defined as the actual yield (what we want to find) divided by the theoretical yield (the maximum possible yield) multiplied by 100%. Mathematically, it can be represented as:

% yield = (actual yield / theoretical yield) * 100

We are given the percent yield as 45.0% and the theoretical yield from step 2 is 1.386 moles. We can rearrange the equation and solve for the actual yield:

actual yield = (% yield/100) * theoretical yield
actual yield = (45.0 / 100) * 1.386
actual yield = 0.6237 moles (rounded to four decimal places)

Therefore, the actual number of moles of HCl gas produced is approximately 0.6237 moles.

Comparing this to the answer choices, we can see that option 3: "0.622 mol" is the closest match.

Well, let's calculate the moles of Cl2:

We have 15.5 L of Cl2 at STP, which means at standard temperature and pressure. At STP, 1 mole of any gas occupies a volume of 22.4 L. So, we can convert the liters of Cl2 to moles by using the following ratio:

1 mole Cl2 / 22.4 L Cl2

15.5 L Cl2 x (1 mole Cl2 / 22.4 L Cl2) = 0.692 mol Cl2

Now, the reaction is 1 mole of Cl2 produces 2 moles of HCl. So, we can use the following ratio to calculate the moles of HCl produced:

2 moles HCl / 1 mole Cl2

0.692 mol Cl2 x (2 moles HCl / 1 mole Cl2) = 1.38 mol HCl

Therefore, the correct answer is option 5: 1.38 mol.

b nh

With gases you may use a shortcut in which L are used directly as mols.

L H2 @ STP and 100% yield = 15.5L Cl2 x (2 mols HCl/1 mol Cl2) = 31 L H2O at 100%. At 45% it is 31 x 0.45 = ? L HCl

0.98