# How many liters are in 16 grams of H_2 at STP?

Jun 16, 2016

$V \cong 1.8 \times {10}^{2} \setminus L$

#### Explanation:

Use the ideal gas equation

$P \cdot V = n \cdot R \cdot T$

Where:

$P \to \text{ is the pressure expressed in } a t m$

$V \to \text{ is the volume occupied by the gas expressed in } L$

$n \to \text{ is the number of moles of the gas}$

$R \to \text{ is the universal gas constant} = 0.0821 \setminus L \cdot a t m \cdot m o {l}^{-} 1 \cdot {K}^{-} 1$

$T \to \text{ is the kelvin temperature }$

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$S . T . P \text{conditions} \implies T = 273 K \mathmr{and} P = 1.00 \setminus a t m$

Rearrange the formula and solve for V.

$V = \frac{n \cdot R \cdot T}{P}$

Find the number of moles of the hydrogen gas present in the 16 grams.

${n}_{{H}_{2}} = 16 \setminus g \setminus {H}_{2} \times \frac{1 \setminus m o l . {H}_{2}}{2.016 \setminus g \setminus {H}_{2}}$

${n}_{{H}_{2}} \cong 7.9 \setminus m o l .$

$V = \frac{7.9 \setminus m o l . \times 0.0821 \setminus L \cdot a t m \cdot m o {l}^{-} 1 \cdot {K}^{-} 1 \times 273 \setminus K}{1.00 \setminus a t m}$

$V = \frac{7.9 \setminus \cancel{m o l .} \times 0.0821 \setminus L \cdot \cancel{a t m} \cdot \cancel{m o {l}^{-} 1} \cdot \cancel{{K}^{-} 1} \times 273 \setminus \cancel{K}}{1.00 \setminus \cancel{a t m}}$

$V \cong 1.8 \times {10}^{2} \setminus L$

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A quick approach

At S.T.P you can use the following formula:

${n}_{{H}_{2}} = \frac{V}{V} _ M$

${n}_{{H}_{2}} \text{ is the number of moles of the gas.}$

$V \text{ is Volume of the gas under S.T.P conditions}$

${V}_{M} \text{ is the molar volume i.e the volume occupied by 1 mole of }$
$\text{any gas under S.T.P conditions, equal to 22.4 L/mol.}$

${n}_{{H}_{2}} = 16 \setminus g \setminus {H}_{2} \times \frac{1 \setminus m o l . {H}_{2}}{2.016 \setminus g \setminus {H}_{2}}$

${n}_{{H}_{2}} \cong 7.9 \setminus m o l .$

$V = {n}_{{H}_{2}} \times {V}_{M}$

$V = 7.9 \setminus m o l \times 22.4 L \cdot m o {l}^{-} 1$

$V \cong 1.8 \times {10}^{2} \setminus L$