Question

When 5.35g of Sodium Nitrate were heated in an open crucible, the mass of oxygen produced was 0.83g. given that the equation for the reaction is:-
2NaNO_3(s) → 2NaNO_2(s) + O_2(g)
Calculate the percentage of Sodium Nitrate that was converted to sodium nitrite. (3marks)
(Na=23, O=16, N = 14)

Answer 1

Moles of oxygen gas. = ^0.83/₃₂
= 0.02594 
Moles of 2NaNO₃ : O₂
                          2 : 1
Moles of NaNO₃ = 2 x 0.02594.
= 0.05188 moles 
Mass of NaNO₃ = 85 x 0.05188
= 4.4098g 
Percentage of NaNO₃ = ^4.098/_5.35 X ^4.098/_5.35 x 100 = 82.45%