A 1.249g mixture containing calcium chloride and sodium chloride reacts with exc
ID: 547330 • Letter: A
Question
A 1.249g mixture containing calcium chloride and sodium chloride reacts with excess sodium carbonate as per equation to precipitate 0.567g of calcium carbonate. a) calculate the moles of calcium carbonate precipitated? b) how many moles of calcium chloride are present in the mixture? c) convert moles of calcium chloride into grams of calcium chloride. d) calculate % calcium chloride in the mixture. A 1.249g mixture containing calcium chloride and sodium chloride reacts with excess sodium carbonate as per equation to precipitate 0.567g of calcium carbonate. a) calculate the moles of calcium carbonate precipitated? b) how many moles of calcium chloride are present in the mixture? c) convert moles of calcium chloride into grams of calcium chloride. d) calculate % calcium chloride in the mixture. a) calculate the moles of calcium carbonate precipitated? b) how many moles of calcium chloride are present in the mixture? c) convert moles of calcium chloride into grams of calcium chloride. d) calculate % calcium chloride in the mixture.Explanation / Answer
a)
mol of CaCO3= mass/MW = 0.567/100 = 0.00567 mol of CaCO3
b)
mol of CaCl2 present
mol of CaCO3 = 0.00567
raito is 1:1 so
0.00567 mol of CaCl2 present
c)
mol of CaCl2 = mol*MW = 0.00567 *110.98 =0.6292566 g of CaCl2
d)
% CaCl2 / sample = 0.6292566 / 1.249*100= 50.38 % is CaCl2
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