Gas formation from reaction #1

The reaction between copper(II)sulfide and nitric acid gives copper ion, sulfur, nitrogen monoxide and water. If 71.5gr of copper(II)sulfide react with 100gr of nitric acid, compute the mass of sulfur and the volume of nitrogen monoxide that form.

SOLUTION:
The reaction is:
$$CuS+HNO_3 \rightarrow Cu^{++}+S+NO+H_2O$$
You can easily balance it:
$$\underset{100gr}{\underbrace{CuS}}+\underset{71.5gr}{\underbrace{{\color{red} 2}H^++{\color{red} 2}NO_3^-}} \rightarrow Cu^{++}+\underset{m?}{\underbrace{S}}+\underset{V?}{\underbrace{{\color{red} 2}NO}}+H_2O$$

We can calculate the number of mol (n) of CuS and $HNO_3$:

$$n_{CuS}=\frac{m_{CuS}}{MW_{CuS}}=\frac{71.5gr}{95.62gr/mol}=0.748mol$$

$$n_{HNO_3}=\frac{m_{HNO_3}}{MW_{HNO_3}}=\frac{100gr}{63gr/mol}=1.587mol$$

For every mol of CuS we obtain one mol of sulfur: $n_{CuS}=n_{S}$. So the mass of sulfur is:
$$m_S=n_S \cdot MW_S=0.748mol \cdot 32.07gr/mol=23.99gr$$

The NO forms from nitric acid. The number of moles of nitric acid is equal to the number of moles of NO: $n_{HNO_3}=n_{NO}$. The volume of one mole of gas is equal to 22.4 liters at standard conditions (0°C, 1atm):
$$V_{NO}=n_{NO} \cdot 22.4L/mol=1.587mol\cdot22.4L/mol=34.55L$$

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PROBLEM 18:
2.0 grams of propane react with 7.0 grams of oxygen; suppose that the reaction is complete, and forms water and CO. Calculate the mass of water and the volume of CO and $O_2$.