Đáp án:
\(1/\\ a,\ V=V_{H_2}=1,344\ lít.\\ m_{\text{dd HCl}}=120\ g.\\ b,\ m_{CuO}=4,8\ g.\\ 2/\\ a,\ V=V_{H_2}=6,72\ lít.\\ b,\ C_{M_{H_2SO_4}}(dư)=\dfrac{0,2}{0,5}=0,4\ M.\\ C_{M_{Al_2(SO_4)_3}}=\dfrac{0,1}{0,5}=0,2\ M.\)
Giải thích các bước giải:
\(1/\\ a,\ PTHH:Mg+2HCl\to MgCl_2+H_2↑\\ n_{Mg}=\dfrac{1,44}{24}=0,06\ mol.\\ Theo\ pt:\ n_{H_2}=n_{Mg}=0,06\ mol.\\ \Rightarrow V=V_{H_2}=0,06\times 22,4=1,344\ lít.\\ Theo\ pt:\ n_{HCl}=2n_{Mg}=0,12\ mol.\\ ⇒m_{HCl}=0,12\times 36,5=4,38\ g.\\ ⇒m_{\text{dd HCl}}=\dfrac{4,38}{3,65\%}=120\ g.\\ b,\ PTHH:CuO+H_2\xrightarrow{t^o} Cu+H_2O\\ Theo\ pt:\ n_{CuO}=n_{H_2}=0,06\ mol.\\ ⇒m_{CuO}=0,06\times 80=4,8\ g.\\ 2/\\ a,\ PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\\ n_{Al}=\dfrac{5,4}{27}=0,2\ mol.\\ n_{H_2SO_4}=0,5\times 1=0,5\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,2}{2}<\dfrac{0,5}{3}\\ ⇒H_2SO_4\ dư.\\ Theo\ pt:\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ ⇒V=V_{H_2}=0,3\times 22,4=6,72\ lít.\\ b,\ n_{H_2SO_4}(dư)=0,5-\dfrac{0,2\times 3}{2}=0,2\ mol.\\ Theo\ pt:\ n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1\ mol.\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=500\ ml=0,5\ lít.\\ ⇒C_{M_{H_2SO_4}}(dư)=\dfrac{0,2}{0,5}=0,4\ M.\\ C_{M_{Al_2(SO_4)_3}}=\dfrac{0,1}{0,5}=0,2\ M.\)
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