Đáp án:
\(\% {m_{Al}} = 32,5\%; \% {m_{Fe}} = 67,5\% \)
\({m_{{H_2}S{O_4}}} = 2,45{\text{ gam}}\)
\({m_{muối}} = 3,23{\text{ gam}}\)
Giải thích các bước giải:
Gọi số mol \(Al;Fe\) lần lượt là \(x;y\)
\( \to 27x + 56y = 0,83\)
Phản ứng xảy ra:
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{3}{2}{n_{Al}} + {n_{Fe}} = 1,5x + y = \frac{{0,56}}{{22,4}} = 0,025{\text{ mol}}\)
Giải được: \(x=y=0,01\)
\( \to {m_{Al}} = 0,01.27 = 0,27{\text{ gam}}\)
\( \to \% {m_{Al}} = \frac{{0,27}}{{0,83}} = 32,5\% \to \% {m_{Fe}} = 67,5\% \)
Ta có:
\({n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,025{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,025.98 = 2,45{\text{ gam}}\)
BTKL:
\({m_{kl}} + {m_{{H_2}S{O_4}}} = {m_{muối}} + {m_{{H_2}}}\)
\( \to 0,83 + 0,025.98 = {m_{muối}} + 0,025.2\)
\( \to {m_{muối}} = 3,23{\text{ gam}}\)
