# Question 50275

Aug 2, 2015

Volume needed: 59 mL.

#### Explanation:

Start from the balanced chemical equation

$M {g}_{\left(s\right)} + \textcolor{red}{2} H C {l}_{\left(a q\right)} \to M g C {l}_{2 \left(a q\right)} + {H}_{2 \left(g\right)} \uparrow$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between hydrochloric acid and hydrogen gas. This means that, for every 2 moles of hydrochloric acid consumed, the reaction will produce 1 mole of hydrogen gas.

This means that you an use the number of moles of gas produced to backtrack and determine the number of moles of hydrochloric acid that reacted.

Since the hydrogen gas is at STP conditions, which imply a pressure of 100kPa and a temperature of 273.15 K, you can use the molar volume of a gas at STP to determine how many moles of hydrogen were produced.

At STP conditions, one mole of any ideal gas occupies exactly 22.7 L. Since your sample occupies 2.00 L, then it follows that you have

2.00color(red)(cancel(color(black)("L"))) * "1 mole"/(22.7color(red)(cancel(color(black)("L")))) = "0.08811 moles" ${H}_{2}$

produced by the reaction. This means that

0.08811color(red)(cancel(color(black)("moles "H_2))) * (color(red)(2)" moles "HCl)/(1color(red)(cancel(color(black)("mole "H_2)))) = "0.1762 moles" $H C l$

reacted with the excess magnesium.

Since a solution's molarity is defined as the number of moles of solute, in this case hydrochloric acid, per liters of solution, it follows that you need

$C = \frac{n}{V} \implies V = \frac{n}{C}$

V = (0.1762color(red)(cancel(color(black)("moles"))))/(3.0color(red)(cancel(color(black)("moles")))/"L") = "0.05873 L"

of solution to deliver that many moles of $H C l$ to th reaction.

Expressed in mililiters and rounded to two sig figs, the number of sig figs you gave for the molarity of the $H C l$ solution, the answer will be

0.05873color(red)(cancel(color(black)("L"))) * "1000 mL"/(1color(red)(cancel(color(black)("L")))) = color(green)("59 mL")#

SIDE NOTE Many textbooks and online sources still list the molar volume of a gas at STP as being equal to 22.4 L.

This value is based on the old definition of STP, which had a pressure of 1 atm and a temperature of 273.15 K. If this is the value you're supposed to use, simply redo the calculations using 22.4 instead of 22.7L.