Đáp án:

\({m_{{H_2}O}} = 3,6{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Ca + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}\)

\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)

Ta có:

\({n_{Ca}} = \frac{{19,5}}{{40}} = 0,4875{\text{ mol = }}{{\text{n}}_{{H_2}}}\)

\({n_{CuO}} = \frac{{16}}{{64 + 16}} = 0,2{\text{ mol < }}{{\text{n}}_{{H_2}}}\)

\( \to {n_{{H_2}O}} = {n_{CuO}} = 0,2{\text{ mol}}\)

\( \to {m_{{H_2}O}} = 0,2.18 = 3,6{\text{ gam}}\)

\(\begin{array}{l}
n_{CaO}=\frac{19,5}{40}=0,4875(mol)\\
n_{Cu}=\frac{16}{80}=0,2(mol)\\
Ca+2HCl\to CaCl_2+H_2\\
Theo\,PT:\,n_{H_2}=n_{Ca}=0,4875(mol)\\
CuO+H_2\xrightarrow{t^o}Cu+H_2O\\
Vi\,n_{CuO}=0,2<0,4875=n_{H_2}\to H_2\,du\\
Theo\,PT:\,n_{H_2O}=n_{Cu}=0,2(mol)\\
\to m_{H_2O}=0,2.18=3,6(g)
\end{array}\)