Calculate the percentage of purity in a sample of impure calcium carbonate, if 2.36g of the sample requires 1.49g of Hydrocholric acid for complete reaction?How do you calculate the percentage of purity? ?
Reaction is CaCO3 + 2HClaq ==%26gt; CaCl2aq + CO2g + H2O
1,49/ 36.5 = 0.0408 mol of HCl
So 0.0204 mol of CaCO3 reacted = 2.04g
sp % purity = 2.04/2.36 = 86,5 %
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