Question

what is the volume of oxygen gas at STP from the decomposition of 1.70 g of sodium nitrate (85.00 g/mol)?

Answer 1

MOlar mass of NaNO₃ = At.mass of Na + At.mass of N + (3xAt.mass of O )

                                  = 23 + 14 + (3x16)

                                  = 85 g/mol

Given mass of NaNO₃ = 1.70 g

So number of moles of NaNO₃ = mass/molar mass

                                            = 1.70 g / 85 (g/mol)

                                            = 0.02 mol

The decomposition of sodium nitrate is : 2 NaNO₃ ----> 2NaNO₂ + O₂

According to the balanced equation ,

2 moles of NaNO₃ produces 1 mole of oxygen

0.02 moles of NaNO₃ produces M mole of oxygen

M = ( 0.02x1) / 2

    = 0.01 moles of Oxygen

Calculation of number of moles of Oxygen :

We know that PV = nRT

Where

T = Temperature = 273 K

P = pressure = 1.0 atm

n = No . of moles = 0.01

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = ?

Plug the values we get V = (nRT) / P

                                      = ( 0.01x0.0821x273) / 1.0

                                      = 0.224 L

Therefore the volume of oxygen gas liberated is 0.224 L = 224 mL