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Consider the chemical equation. CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to produce 21.2 g of NaCl?
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CuCL2 = 63.55 + 2*35.45 = 134.45 2 Na Cl = 2(23 + 35.450 = 116.90
so 134.45 of CuCl2 should produce 116.90 gm of NaCL if the yield is 100% 31g should produce 116.9 * 31 -------- g of NACL = X g 134.45 % yield will 21.2 * 100 --------- X
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