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Câu hỏi: Cho $a$ gam Mg vào 100ml dung dịch Al2(SO4)3 1M và CuSO4 3M thu được 21,9 gam hỗn hợp chất rắn gồm hai kim loại. Giá trị của a là
A. 14,4.
B. 21,6.
C. 13,4.
D. 10,8.
A. 14,4.
B. 21,6.
C. 13,4.
D. 10,8.
$a\left( g \right)Mg\xrightarrow[CuS{{O}_{4}}:0,3\left( mol \right)]{A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}:0,1\left( mol \right)}21,9\left( g \right)\left\{ \begin{aligned}
& Al \\
& Cu \\
\end{aligned} \right.$
$\xrightarrow{BTNT\left( Al \right)}{{n}_{Cu}}=2{{n}_{CuS{{O}_{4}}}}=0,3\left( mol \right)\to {{m}_{Cu}}=19,2\left( g \right)$
$\to {{m}_{Al}}=21,9-19,2=2,7\left( g \right)\to {{n}_{Al}}=0,1\left( mol \right)\to {{n}_{A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\left( p.u \right)}}=0,05\left( mol \right)$
BT $e:{{n}_{Mg}}=\dfrac{6{{n}_{A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\left( p.u \right)}}+2{{n}_{CuS{{O}_{4}}\left( p.u \right)}}}{2}=\dfrac{6.0,05+2.0,3}{2}=0,45\left( mol \right)$
$\to a=10,8\left( g \right)$
& Al \\
& Cu \\
\end{aligned} \right.$
$\xrightarrow{BTNT\left( Al \right)}{{n}_{Cu}}=2{{n}_{CuS{{O}_{4}}}}=0,3\left( mol \right)\to {{m}_{Cu}}=19,2\left( g \right)$
$\to {{m}_{Al}}=21,9-19,2=2,7\left( g \right)\to {{n}_{Al}}=0,1\left( mol \right)\to {{n}_{A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\left( p.u \right)}}=0,05\left( mol \right)$
BT $e:{{n}_{Mg}}=\dfrac{6{{n}_{A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\left( p.u \right)}}+2{{n}_{CuS{{O}_{4}}\left( p.u \right)}}}{2}=\dfrac{6.0,05+2.0,3}{2}=0,45\left( mol \right)$
$\to a=10,8\left( g \right)$
Đáp án D.