# What is the volume of H_2 generated by reacting 4.33g of Zn with excess H_2SO_4 at 38°C and 760 torr?

Dec 29, 2015

$\text{1.7 L}$

#### Explanation:

Zinc metal will react with dilute sulfuric acid to produce zinc sulfate, ${\text{ZnSO}}_{4}$, and hydrogen gas, ${\text{H}}_{2}$, which will bubble out of solution.

The balanced chemical equation for this single replacement reaction looks like this

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

Notice that you have a $1 : 1$ mole ratio between zinc metal and hydrogen gas. This tells you that the reaction will always produce the same number of moles of hydrogen gas as you have moles of zinc metal that take part in the reaction.

In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.

Use zinc metal's molar mass to determine how many moles you have in that $\text{4.33-g}$ sample

4.33 color(red)(cancel(color(black)("g"))) * "1 mole Zn"/(65.38color(red)(cancel(color(black)("g")))) = "0.06623 moles Zn"

This means that the reaction produced $0.06623$ moles of hydrogen gas.

Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this

$\textcolor{b l u e}{P V = n R T} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, always expressed in Kelvin!

It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.

This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.

So, rearrange the ideal gas law equation to solve for $V$, the volume of the gas

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

$V = \left(0.06623 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("moles"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 38)color(red)(cancel(color(black)("K"))))/(760/760color(red)(cancel(color(black)("atm}}}}\right)$

$V = \text{1.692 L}$

Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be

$V = \textcolor{g r e e n}{\text{1.7 L}}$

Dec 29, 2015

${\text{V"= 1.690"dm}}^{3}$

## State the Equation

Zinc's only common oxidation state is $\text{2+}$. When it reacts with sulfuric acid, the result is the following equation:

${\text{Zn"(s) + "H"_2"SO"_4(aq) -> "ZnSO"_4 (aq) + "H}}_{2} \left(g\right)$

## Calculating Moles of ${\text{H}}_{2}$

First, calculate the moles of $\text{Zn}$ reacting:

${\text{mol" = "m"/"M}}_{r}$
"mol"("Zn") = (4.33"g")/65.4
$\text{mol"("Zn") = 0.06620795107" moles}$

Next, calculate the moles of ${\text{H}}_{2}$ produced using the mole ratio:

$\text{mol"("H"_2) = "mol"("Zn") = 0.06620795107" moles}$

## Calculating Volume of ${\text{H}}_{2}$

In order to find the volume of space occupied by the calculated number of moles of hydrogen gas at the stated conditions, we must consult the ideal gas equation:

$\text{pV " = " nRT}$

Where:

• $\text{p}$ is pressure in pascals ($\text{Pa}$)
• $\text{V}$ is volume in cubic metres ( ${\text{m}}^{3}$)
• $\text{n}$ is the number of moles
• $\text{R}$ is the gas constant = $8.314$
• $\text{T}$ is the temperature in Kelvin ($\text{K}$)

First, convert the given data into workable units:

• $1 \text{torr" ~~ 133.322"Pa", :. 760"torr" = 101325"Pa}$
• ${1}^{\circ} \text{C" = 274.15"K", :. 38^@"C" = 311.15"K}$

Next, rearrange the equation to solve for volume:

$\text{pV "=" nRT " => " V "=" ""nRT"/"p}$

Finally, substitute in your values and solve the equation:

$\text{V "=" } \frac{0.06620795107 \cdot 8.314 \cdot 311.15}{101325}$
$\text{V "=" "1.69 * 10^(-3)"m"^3" to 3 significant figures}$

Another useful unit that is probably suitable to use in this situation is the cubic decimetre ( ${\text{dm}}^{3}$ ):

${\text{V "=" "1.69 * 10^(-3)"m"^3 * 1000 = 1.690"dm}}^{3}$