Given balanced equation C3H3+5O2----->3CO2+4H2O
First find the moles of C3H3, given mass of C3H3=500 g and molar mass=39 g/mol
So mol of C3H3=mass/molar mass=500 g/39 g/mol=12.82 mols C3H3
Now from the equation mol ratio between C3H3 and CO2 is 1:3
Then moles of CO2=3xmoles of C3H3=3x12.82 mol=38.46 mol CO2.
Now multiply with molar mass of CO2 we will gwt theoretical yield.
mass of CO2 produced=38.46 molx44 g/mol=1692.30 g
Given experimental yield of CO2=1000 g.
So % yield= (experimental yield/theoretical yield)x100
% yield=(1000 g/1692.3 g)100=59.09 %.
Shortcut is (500g/39g)x(3 mol CO2)x44 g/mol=1692.3 g
% yield=(1000/1692.3)x100=59.09%.
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