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?

Question

3420 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-12T01:51:26

$1.67\times10^-1color(white)(l)"mol"$

The Reaction:

$Mg(s)+2HCl(aq)\rightarrowMgCl_2(aq)+H_2(g)$

Data given:

Pressure $P="817.36 mmHg"$
Volume $V="4.03 L"$
Temperature $T="303 K"$

Work :

$P_"total" = P_"H₂" + P_"H₂O"$

At 303 K, $P_"H₂O" = "33.4 mmHg"$

∴ $P_"H₂" = P_"total" - P_"H₂O" = "817.36 mmHg" - "33.4 mmHg" = "783.96 mmHg" = "1.0315 atm"$

Apply the Ideal Gas Law $PV=nRT$ to get $n = \frac(PV)(RT)$

The constant $R = "0.0821 L·atm·K"^"-1""mol"^"-1"$

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