From 20 kg of CaCO3 limestone (3-index), containing 4% of impurities, 12 kg of Ca (OH) 2 were obtained (twice). What percentage is this compared to the theoretical yield?
From 1 mole of limestone it is possible to synthesize 1 mole of slaked lime.
Let’s calculate the available chemical amount of limestone.
M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;
N CaCO3 = 20,000 x 0.96 / 100 = 192 mol;
The same amount of slaked lime can be synthesized.
Let’s calculate the theoretical chemical amount and weight of slaked lime.
M Ca (OH) 2 = 40 + 16 x 2 + 2 = 74 grams / mol;
m Ca (OH) 2 = 192 x 74 = 14 208 grams;
The reaction yield is 12,000 / 14.208 = 0.845 = 84.5%;
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