A. 16,20
B. 42,12
C. 32,40
D. 48,60
B
\(\begin{array}{l}
{n_{Fe}} = 0,15\,mol;\,{n_{AgN{O_3}}} = 0,39\,mol\\
Fe + 2AgN{O_3} \to Fe{\left( {N{O_3}} \right)_2} + 2Ag\\
0,15\,\,\,\,\,\,\,\,\,0,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\\
Fe{\left( {N{O_3}} \right)_2} + AgN{O_3} \to Fe{\left( {N{O_3}} \right)_3} + Ag\\
0,09\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,09\\
\Rightarrow {n_{Ag}} = 0,39\,mol\\
\Rightarrow {m_{Ag}} = 42,12g
\end{array}\)
→ Đáp án B
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