Đáp án:

\(\begin{array}{l}
85)\quad \rm D.\, 64\, gam\\
86)\quad \rm C.\, 80\%
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
85)\quad \rm CaC_2 + 2H_2O \longrightarrow C_2H_2 + Ca(OH)_2\\
\rm n_{C_2H_2} = \dfrac{17,92}{22,4} = 0,8\, mol\\
\rm n_{CaC_2} = n_{C_2H_2} = 0,8\, mol\\
\rm m_{CaC_2} = 0,8\times 64= 51,2\, gam\\
\rm m_{đất\,đèn}=\dfrac{51,2\times 100}{80} = 64\, gam\\
86)\quad \rm CaC_2 + 2H_2O \longrightarrow C_2H_2 + Ca(OH)_2\\
\rm n_{C_2H_2} = \dfrac{0,224}{22,4} = 0,01\, mol\\
\rm n_{CaC_2} = n_{C_2H_2} = 0,01\, mol\\
\rm m_{CaC_2} = 0,01\times 64 = 0,64\, gam\\
\rm \%m_{CaC_2} = \dfrac{0,64}{0,8}\times 100\% = 80\%
\end{array}\)

Đáp án:

 85 D

86 C

Giải thích các bước giải:

\(\begin{array}{l}
85)\\
Ca{C_2} + 2{H_2}O \to {C_2}{H_2} + Ca{(OH)_2}\\
{n_{{C_2}{H_2}}} = \dfrac{{17,92}}{{22,4}} = 0,8\,mol\\
{n_{Ca{C_2}}} = {n_{{C_2}{H_2}}} = 0,8\,mol\\
{m_{Ca{C_2}}} = 0,8 \times 64 = 51,2g\\
{m_{\text{ đất đèn }}} = \dfrac{{51,2}}{{80\% }} = 64g\\
86)\\
Ca{C_2} + 2{H_2}O \to {C_2}{H_2} + Ca{(OH)_2}\\
{n_{{C_2}{H_2}}} = \dfrac{{0,224}}{{22,4}} = 0,01mol\\
{n_{Ca{C_2}}} = {n_{{C_2}{H_2}}} = 0,01\,mol\\
{m_{Ca{C_2}}} = 0,01 \times 64 = 0,64g\\
\% {m_{Ca{C_2}}} = \dfrac{{0,64}}{{0,8}} \times 100\%  = 80\% 
\end{array}\)