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How many moles of ZnCl2 will be produced from 23.0 g of Zn,
assuming CuCl2 is available in excess?
Zn + CuCl2 → ZnCl2 + Cu
-----------------------------------------
Zn의 몰질량 = 65.38 g/mol
23.0 g / (65.38 g/mol) = 0.351790 mol Zn
( 계산 설명 http://ywpop.tistory.com/7738 )
Zn : ZnCl2 = 1 : 1 계수비(몰수비) 이므로,
0.351790 mol Zn = 0.351790 mol ZnCl2
답: 0.352
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