# Question #3f409

Jun 27, 2014

The mass of KNO₂ is 21 g; the volume of O₂ is 3.0 L.

Part (1) is a stoichiometry problem.

You must make the following conversions:

mass of KNO₃ → moles of KNO₃ → moles of KNO₂ → mass of KNO₂

The balanced equation is

2KNO₃ → 2KNO₂ + O₂

The molar mass of KNO₃ is (39.098 + 14.007 + 3× 15.999) g = 101.102 g

25 g KNO₃ × $\left(1 \text{mol KNO₃")/(101.102"g KNO₃}\right)$ = 0.2473 mol KNO₃

(2 significant figures + 2 guard digits).

The molar mass of KNO₂ is (39.098 + 14.007 + 2 × 15.999) g = 85.103 g

0.2473 mol KNO₃ × $\left(2 \text{mol KNO₂")/(2"mol KNO₃")× (85.103"g KNO₂")/(1"mol KNO₂}\right)$ = 21 g KNO₂

(2 significant figures)

Part (2) is a gas stoichiometry problem.

You must convert

moles of KNO₃ → moles of O₂ → volume of O₂

0.2473 mol KNO₃ × $\left(1 \text{mol O₂")/(2"mol KNO₃}\right)$ = 0.1236 mol O₂

Room temperature and pressure are not well defined. I will assume that RTP = 1 atm and 25 °C.

We can use the Ideal Gas Law to calculate the volume of O₂.

$P V = n R T$

$T$ = (25 + 273.15) K = 298.15 K

$V = \frac{n R T}{P} = \left(0.1236 \text{mol" × 0.08206"L·atm·K⁻¹mol⁻¹" × 298.15 "K")/(1"atm}\right)$ = 3.0 L

Hope this helps.