Question

A 0.4230 g sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to 0.1064 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer 1

first write balanced equation

2NaNO₃ ===> 2NaNO₂ + O₂

no of moles of NaNO2 = weight of NaNO2 / molar mass of NaNO2

= 0.1064 / 69

= 0.00154 moles

from the balanced equation it clear that 2 moles of NaNO2 will get 2 moles of NaNO3

that means 0.00154 mol will get from 0.00154 moles of NaNO3

now calculate the weight of NaNO3 from the moles

weight = no of moles of NaNO3 x molar mass of NaNO3

= 0.00154 x 84.977

= 0.131 grams

this is the original weight of NaNO3

% NaNO3 in original sample

= 0.131 / 0.423 x 100

= 31 %