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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:-
2NaNO3(s) → 2NaNO2(s) + O2(g)
Calculate the percentage of Sodium Nitrate that was converted to sodium nitrite. (3marks)
(Na=23, O=16, N = 14)

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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%

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