Propylene, C3H6, reacts with hydrogen under pressure to give propane, C3H8: C3H6(g) + H2(g) -----> C3H8(g) How many liters of hydrogen ( at 740 torr and 24 degree C) react with 18.0 g of propylene?

Mar 25, 2015

You would need $\text{11 L}$ of oxygen to react with that much propylene under those conditions for temperature and pressure.

${C}_{3} {H}_{6 \left(g\right)} + {H}_{2 \left(g\right)} \to {C}_{3} {H}_{8 \left(g\right)}$

Notice that you have a $\text{1:1}$ mole ratio between propylene and hydrogen, which means that the number of moles of hydrogen you need is equal to the number of moles of propylene that react.

Use propylene's molar mass to determine how many moles you have

$\text{18.0"cancel("g") * "1 mole propylene"/("42.08"cancel("g")) = "0.4278 moles propylene}$

Automatically, you have

$\text{0.4278"cancel("moles propylene") * "1 mole hydrogen"/cancel("1 mole propylene") = "0.4278 moles hydrogen}$

Now use the ideal gas law equation to solve for the volume of hydrogen - don't forget to use Kelvin and atm

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

$V = \left(0.4278 \cancel{\text{moles") * 0.082(cancel("atm") * "L")/(cancel("mol") * cancel("K")) * (273.15 + 24)cancel("K"))/(740/760cancel("atm}}\right)$

$V = \text{10.705 L}$

Rounded to two sig figs, the number of sig figs given for 24 degrees Celsius, the answer will be

$V = \textcolor{red}{\text{11 L}}$