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what is the percent yield of oxygen in a sample of 115g of oxygen produced by heating 400g of potassium chlorate

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The % yield of oxygen in a sample of 115g of oxygen produced by heating 400g of potassium chlorate is 73.37%

calculation

% yield = actual yield/ theoretical yield x100

the actual yield = 115 g

f
ind the theoretical yield by

first write the equation for decomposition of KClO3
that is 2KClO3 = 2KCl +3 O2

then find the moles of KClO3 = mass/molar mass

=400g/122.5 g/mol = 3.265 moles
by use of mole ratio between KClO3 to O2 which is 2:3 the moles of O2= 3.265 x3/2 =4.898 moles

the theoretical mass of O2 =moles of O2 x molar mass oF O2

=4.898 x 32 g/mol = 156.74 grams
% yield is therefore =115/ 156.74 x100 = 73.37%

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