Question

How many grams of NH4NO3 are needed to produce 17.4 L of oxygen?

Answer 1

$$2NH_4NO_3(s)\rightarrow2N_2(g)+O_2(g)+4H_2O(g)$$ $$n=\frac{V}{V_m}$$

$$n(O_2)=\frac{V}{V_m}=\frac{17.4L}{22.4\frac{L}{mol}}=0.777 mol$$

According to equation $n(NH_4NO_3):n(O_2)=2:1$, then $n(NH_4NO_3)=n(O_2)\times2=0.777mol\times2=1.55 mol$

$$m=M\times n$$

$$m(NH_4NO_3)=80.04\frac{g}{mol}\times 1.55mol= 124g$$