# Question 7a784

Dec 7, 2015

$\text{170 mL}$

#### Explanation:

Your strategy here will be to

• write a balanced chemical equation for this single replacement reaction

• use the molar volume of a gas at STP to find the number of moles of hydrogen gas present in that volume

• use the mole ratio that exists between sulfuric acid and hydrogen gas to find the number of moles of the former needed to produce that many moles of the latter

• use the sulfuric acid solution's molarity to find the volume that would contain that many moles of sulfuric acid

So, the balanced chemical equation for this reaction looks like this

$2 {\text{Al"_text((s]) + 3"H"_2"SO"_text(4(aq]) -> "Al"_2("SO"_4)_text(3(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

As you know, STP conditions are characterized by a pressure of $\text{100.0 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions, one mole of any ideal gas occupies exactly $\text{22.71 L}$ - this is the molar volume of a gas at STP.

If one mole of hydrogen gas would occupy $\text{22.71 L}$ at STP, it follows that your volume would contain

15.0 color(red)(cancel(color(black)("L"))) * "1 mole H"_2/(22.71 color(red)(cancel(color(black)("L")))) = "0.6605 moles H"_2

Now look at the balanced chemical equation. Notice that $3$ moles of sulfuric acid will produce $3$ moles of hydrogen gas $\to$ a $1 : 1$ mole ratio exists between the two chemical species.

Simply put, the number of moles of sulfuric acid that took part in the reaction is equal to the number of moles of hydrogen gas produced by the reaction.

${n}_{{H}_{2} S {O}_{4}} = {n}_{{H}_{2}} = \text{0.6605 moles}$

Finally, since molarity is defined as moles of solute per liters of solution, you can say that

$\textcolor{b l u e}{c = \frac{n}{V} \implies V = \frac{n}{c}}$

The volume of $\text{4.0 M}$ sulfuric acid solution that would contain $0.6605$ moles of sulfuric acid will be equal to

V = ( 0.6605 color(red)(cancel(color(black)("moles"))))/(4.0 color(red)(cancel(color(black)("moles")))/"L") = "0.165 L"#

Expressed in milliliters and rounded to two sig figs, the number of sig figs you have for the molarity of the sulfuric acid solution, the answer will be

$V = \textcolor{g r e e n}{\text{170 mL}}$

SIDE NOTE Many textbooks and online sources still use the old definition of STP conditions, at which pressure is $\text{1 atm}$ and temperature is ${0}^{\circ} \text{C}$.

Under these conditions, one mole of any ideal gas occupies $\text{22.4 L}$. If this is the value you're supposed to use for the molar volume of a gas at STP, simply redo the calculations with $\text{22.4 L}$ instead of $\text{22.71 L}$.