Question

What mass of potassium nitrate is needed to generate 157.0 L of gas, composed of 105.0 L of N2 and 52.0 L of O2 at 0.920 atm and 293 K, using these two reactions?

Answer 1

First, figure out how moles there are in 105 L and 52 L:

$$PV=nRT,\\\space\\ n_{N_2}=\frac{PV}{RT}=4.018\text{ mol},\\\space\\ n_{O_2}=\frac{PV}{RT}=1.990\text{ mol}.$$

Write the reactions and substitute the products from nitrite to make it one 'combined' reaction useful for calculations:

$$2\text{KNO}_3\rightarrow2\text{KNO}_2+\text{O}_2,\\ 4\text{KNO}_2\rightarrow2\text{K}_2\text{O}+3\text{O}_2+2\text{N}_2.$$

Algebraical transformations give us

$$4\text{KNO}_3\rightarrow2\text{K}_2\text{O}+5\text{O}_2+2\text{N}_2.$$

As we see from the reaction, we need two times more nitrate than nitrogen to produce 105 L:

$$n_{nitrate/N_2}=2n_{N_2}=8.036\text{ mol}.$$

Also, 4 molecules of nitrate produce 5 molecules of oxygen:

$$n_{nitrate/O_2}=\frac{4}{5}n_{O_2}=2.488\text{ mol}.$$

The total amount of quantity of potassium nitrate:

$$n_{nitrate}=8.036+2.488=10.52\text{ mol}.$$

What mass of nitrate is required? Multiply it by its molar mass:

$$m_{nitrate}=n_{nitrate}\Mu_{nitrate}=\\=10.52\cdot(39+14+16\cdot3)=1063\text{ g}.$$