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Đáp án + Giải thích các bước giải:

Câu 7:

$\rm 2Na+2H_2O\to 2NaOH+H_2\\ n_{Na}=\dfrac{4,6}{23}=0,2(mol)\\ \xrightarrow{Theo\ pt:} \begin{cases} \rm n_{NaOH}=n_{Na}=0,2(mol)\\\rm n_{H_2}=\dfrac{1}{2}n_{Na}=0,1(mol) \end{cases}\to \begin{cases} \rm m_{NaOH}=0,2.40=8(g)\\\rm V_{H_2}=0,1.22,4=2,24(l) \end{cases}$

Câu 8:

Bảo toàn khối lượng:

$\rm m_{kl}+m_{O_2}=m_{oxide}\\ \to m_{O_2}=10,9-8,5=2,4(g)\\ n_{O_2}=\dfrac{2,4}{32}=0,075(mol)\\ V_{kk}=\dfrac{V_{O_2}}{20\%}=\dfrac{0,075.22,4}{20\%}=8,4(l)$

Câu 9:

Bảo toàn khối lượng:

$\rm m_{pk}+m_{O_2}=m_{oxide}\\ \to m_{O_2}=46,8-21,2=25,6(g)\\ n_{O_2}=\dfrac{25,6}{32}=0,8(mol)\\ V_{kk}=\dfrac{V_{O_2}}{20\%}=\dfrac{0,8.22,4}{20\%}=89,6(l)$

Câu 10:

$\rm a)\\ CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\\ n_{CaCO_3}=\dfrac{20}{100}=0,2(mol)\\ n_{HCl}=2n_{CaCO_3}=0,4(mol)\\ m_{HCl}=0,4.36,5=14,6(g)\\ b)\\ n_{CaCl_2}=n_{CaCO_3}=0,2(mol)\\ m_{CaCl_2}=0,2.111=22,2(g)\\ c)\\ n_{CO_2}=n_{CaCO_3}=0,2(mol)\\ V_{CO_2}=0,2.22,4=4,48(l)$