Question e70cc

Aug 13, 2015

Volume of sulfur dioxide produced: 1060 L.

Explanation:

Copper (II) sulfide, $\text{Cu"_2"S}$, will react with heated oxygen gas to produce copper metal and sulfur dioxide, ${\text{SO}}_{2}$, according to the following balanced chemical equation

${\text{Cu"_2S_text((s]) + "O"_(2(g)) -> color(blue)(2)"Cu"_text((s]) + "SO}}_{\textrm{2 \left(g\right]}}$

Notice that you have a $\textcolor{b l u e}{2} : 1$ mole ratio between copper metal and sulfur dioxide.

This means that the reaction will produce half as many moles of sulfur dioxide than it will of copper metal.

You can use the mass of copper to determine how many moles of copper were produced - use copper's molar mass

5.45color(red)(cancel(color(black)("kg"))) * ("1000" color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole Cu"/(63.55color(red)(cancel(color(black)("g")))) = "85.76 moles Cu"

This means that the reaction also produced

85.76color(red)(cancel(color(black)("moles Cu"))) * ("1 mole SO"""_2)/(color(blue)(2)color(red)(cancel(color(black)("moles Cu")))) = "42.88 moles SO"""_2#

To determine what volume this many moles of sulfur dioxide would occupy, use the ideal gas law equation

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

In your case, you have

${V}_{S {O}_{2}} = \left(42.88 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("moles"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(100/101color(red)(cancel(color(black)("atm}}}}\right)$

${V}_{S {O}_{2}} = \text{1058.8 L}$

SIDE NOTE Do not forget to convert temperature to Kelvin and pressure to atm!

I'll leave the answer rounded to three sig figs, the number of sig figs you gave for the mass of copper, despite the fact that you only gave one sig fig for the pressure at which the gas is collected

${V}_{S {O}_{2}} = \textcolor{g r e e n}{\text{1060 L}}$