Question #f51ce

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Question #f51ce

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Answer 1 · 2017-09-11T00:53:19

At STP, the volume of chlorine used is 90.9 L.

Explanation:

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

$M_{r} : m m m m m m m 36.46$ $M_r:color(white)(mmmmmmm) 36.46$
$m m m H_{2} + {Cl}_{2} \to 2HCl$ $color(white)(mmm)"H"_2 + "Cl"_2 → "2HCl"$

2. Convert grams of $HCl$ $"HCl"$ to moles of $HCl$ $"HCl"$

$Moles of HCl = 292 g HCl \times \frac{1 mol HCl}{36.46 g HCl} = 8.009 mol HCl$ $"Moles of HCl" = 292 color(red)(cancel(color(black)("g HCl"))) × ("1 mol HCl")/(36.46 color(red)(cancel(color(black)("g HCl")))) = "8.009 mol HCl"$

Step 3. Convert moles of $HCl$ $"HCl"$ to moles of ${Cl}_{2}$ $"Cl"_2$

${Moles of Cl}_{2} = 8.009 mol HCl \times \frac{{1 mol Cl}_{2}}{2 mol HCl} = {4.004 mol Cl}_{2}$ $"Moles of Cl"_2 = 8.009color(red)(cancel(color(black)("mol HCl"))) × ("1 mol Cl"_2)/(2 color(red)(cancel(color(black)("mol HCl")))) = "4.004 mol Cl"_2$

4. Calculate the volume of ${Cl}_{2}$ $"Cl"_2$

At STP (1 bar and 0 °C), the molar volume of a gas is 22.71 L.

∴ $V = 4.004 mol \times \frac{22.71 L}{1 mol} = 90.9 L$ $V = 4.004 color(red)(cancel(color(black)("mol"))) × "22.71 L"/(1 color(red)(cancel(color(black)("mol")))) = "90.9 L"$

The volume of ${Cl}_{2}$ $"Cl"_2$ used is 90.9 L.

Here's a useful video on mass-volume conversions.

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