Đáp án:

$Na_{2}O + H_{2}O→2NaOH$

$CuO + H_{2}O→$ không xảy ra phản ứng

1) $m_{hh}=m_{Na_{2}O} + m_{CuO}$

⇒ $m_{Na_{2}O}=m_{hh} - m_{CuO}=21,95 - 8= 13,95 (g)$

⇒ $\begin{cases} \text{%}m_{Na_{2}O}=\dfrac{m_{Na_{2}O}}{m_{hh}} × 100\text{%}=\dfrac{13,95}{21,95} × 100\text{%}= 63,55\text{%}\\\text{%}m_{CuO}=100 - 63,55= 36,45\text{%} \end{cases}$

2) $n_{Na_{2}O}=\dfrac{m}{M}=\dfrac{13,95}{62}= 0,225 (mol)$

$m_{\text{dd sau pứ}}=13,95 + 100= 113,95 (g)$

Theo PT: $n_{NaOH}=2n_{Na_{2}O}=2 × 0,225= 0,45 (mol)$

⇒ $m_{NaOH}=n × M=0,45 × 40= 18 (g)$

⇒ $\text{C%NaOH=$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{18}{113,95}$ × 100%=  15,8%}$

\(\begin{array}{l}
1)\\
Na_2O+H_2O\to 2NaOH\\
\to m_{CuO}=8(g)\\
\to \%m_{CuO}=\frac{8}{21,95}.100\%≈36,45\%\\
\to \%m_{Na_2O}=100-36,45=63,55\%\\
2)\\
n_{Na_2O}=\frac{21,95-8}{62}=0,225(mol)\\
\to n_{NaOH}=2n_{Na_2O}=0,45(mol)\\
m_{dd\,spu}=21,95-8+100=113,95(g)\\
\to C\%_{NaOH}=\frac{0,45.40}{113,95}.100\%≈15,8\%
\end{array}\)