Đề thi thử chuyên nguyễn huệ lần 3

Z

zaichipa2

cau 8 ta co . cus04+2hcl ->cu+cl2+h2s04 ta co ncus04=0,2 mol so mol cua hcl = 0, 04 mol, nh vay cuos4 con du= 0,01-0,02 =0,08 mol. cus04 + h20 ->cu +h2s04+ 1/2 o2. khi do so mol cua khi = 0,04+0,02 =0,06mol, v=1.344 lit :D
 
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