Bài làm :
`n_(CuO)=16/80=0,2 \ (mol)`
a,
$2Cu+O_2 \ \xrightarrow{t^o} \ 2CuO$
b,
Theo PT
`n_(O_2)=1/2 n_(CuO)=1/2 . 0,2=0,1 \ (mol)`
$\rm V_{O_2 \ (đktc)}=0,1.22,4=2,24 \ (l)$
c,
`n_(Cu)=n_(CuO)=0,2 \ (mol)`
`->m_(Cu)=0,2.64=12,8 \ (g)`
Đáp án:
Giải thích các bước giải:
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