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標題:
form.4 chemistry
發問:
5.00g of an impure sample of potassium carbonate(K2CO3)required 34.2cm^3 of 2.00M hydrochloric acid for complete reaction. What is the percentage purity of the carbonate? (Assume that the impurities in the sample do not react with hydrochloric acid.)
最佳解答:
First,you should write the chemical equation: K2CO3(aq) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l) the no. mole of the hydrochloric acid = MV ???????????????= (2.00)(34.2 / 1000) ???????????????= 0.0684 mol By the chemical equation,we know that 2 moles HCl react with 1 moles K2CO3 ∴the no. mole of the K2CO3 = (1/2)(0.0684) ????????????= 0.0342 mol the molar mass of K2CO3 = 2(39.0) + 12.0 + 3(16.0) ?????????? = 138 g mol-1 ∴the no. mole of the K2CO3 = Mass / the molar mass 0.0342 = Mass of K2CO3 / 138 Mass of K2CO3 = 4.7196g ∴the percentage purity of the carbonate = (4. 7196 / 5.00) X 100% = 94.392%
form.4 chemistry
發問:
5.00g of an impure sample of potassium carbonate(K2CO3)required 34.2cm^3 of 2.00M hydrochloric acid for complete reaction. What is the percentage purity of the carbonate? (Assume that the impurities in the sample do not react with hydrochloric acid.)
最佳解答:
First,you should write the chemical equation: K2CO3(aq) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l) the no. mole of the hydrochloric acid = MV ???????????????= (2.00)(34.2 / 1000) ???????????????= 0.0684 mol By the chemical equation,we know that 2 moles HCl react with 1 moles K2CO3 ∴the no. mole of the K2CO3 = (1/2)(0.0684) ????????????= 0.0342 mol the molar mass of K2CO3 = 2(39.0) + 12.0 + 3(16.0) ?????????? = 138 g mol-1 ∴the no. mole of the K2CO3 = Mass / the molar mass 0.0342 = Mass of K2CO3 / 138 Mass of K2CO3 = 4.7196g ∴the percentage purity of the carbonate = (4. 7196 / 5.00) X 100% = 94.392%
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