# If 5.00 L of hydrogen gas measured at 20.0*C and 80.1 kPa is burned in an excess of oxygen, what mass under the same conditions, would be consumed?

Jul 20, 2018

The mass of $\text{H"_2}$ consumed is $\text{0.331 g H"_2}$.

#### Explanation:

Use the equation for the ideal gas law to find the moles of hydrogen gas. Then determine the mass by multiplying the moles by its molar mass.

Ideal gas law

$P V = n R T$

Known

$P = \text{80.1 kPa}$

$V = \text{5.00 L}$

$R = {\text{8.31447 L kPa K"^(-1) "mol}}^{- 1}$

$T = \text{20.0"^@"C + 273.15"="293.2 K}$

Unknown

moles, $n$

To determine moles $\text{H"_2}$, rearrange the equation to isolate $n$. Plug in the known values and solve.

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

n=(80.1color(red)cancel(color(black)("kPa"))xx5.00color(red)cancel(color(black)("L")))/(8.31447color(red)cancel(color(black)("L")) color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx293.2color(red)cancel(color(black)("K")))="0.164 mol H"_2"

To determine the mass of $\text{H"_2}$ by multiplying the moles by its molar mass $\left(\text{2.016 g/mol}\right)$.

0.164color(red)cancel(color(black)("mol H"_2))xx(2.016"g H"_2)/(1color(red)cancel(color(black)("mol H"_2)))="0.331 g H"_2"