# The reaction of magnesium with hydrochloric acid at 30 °C and 817.36 mmHg produced 4.03 L of hydrogen that was collected over water. The vapour pressure of water at 30 °C is 33.4 mmHg. How many moles of hydrogen were formed?

##### 1 Answer
Nov 12, 2016

$1.67 \setminus \times {10}^{-} 1 \textcolor{w h i t e}{l} \text{mol}$

#### Explanation:

The Reaction:

$M g \left(s\right) + 2 H C l \left(a q\right) \setminus \rightarrow M g C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right)$

Data given:

Pressure $P = \text{817.36 mmHg}$
Volume $V = \text{4.03 L}$
Temperature $T = \text{303 K}$

Work :

${P}_{\text{total" = P_"H₂" + P_"H₂O}}$

At 303 K, ${P}_{\text{H₂O" = "33.4 mmHg}}$

${P}_{\text{H₂" = P_"total" - P_"H₂O" = "817.36 mmHg" - "33.4 mmHg" = "783.96 mmHg" = "1.0315 atm}}$

Apply the Ideal Gas Law $P V = n R T$ to get $n = \setminus \frac{P V}{R T}$

The constant $R = \text{0.0821 L·atm·K"^"-1""mol"^"-1}$

\frac("1.0315 atm"\times"4.03 L")("0.0821 L·atm·K"^"-1""mol"^"-1"\times"303 K") = 1.67\times10^-1color(white)(l)"mol"