Question

Giúp với ạ

Answer 1

$a\big)$

$n_{Cu}=\frac{6,4}{64}=0,1(mol)$

$2Cu+O_2\xrightarrow{t^o}2CuO$

Theo PT: $n_{CuO}=n_{Cu}=0,1(mol)$

$\to m_{CuO}=0,1\times 80=8(g)$

$b\big)$

Theo PT: $n_{O_2}=\frac{1}{2}n_{Cu}=0,05(mol)$

$\to V_{O_2(đktc)}=0,05\times 22,4=1,12(l)$

$\to V_{\rm không\,khí}=\frac{1,12}{20\%}=5,6(l)$