Bài 9:

$PTHH:2Na+H_2SO_4\to Na_2SO_4+H_2O\\ n_{Na}=\dfrac{23}{23}=1(mol)\\ \Rightarrow n_{H_2}=n_{Na_2SO_4}=\dfrac{1}{2}n_{Na}=0,5(mol)\\ \Rightarrow V_{H_2}=0,5.22,4=11,2(l);m_{Na_2SO_4}=0,5.142=71(g)$

Bài 10:

$PTHH:Fe+2HCl\to FeCl_2+H_2\\ H_2+S\xrightarrow{t^o}H_2S\\ n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ \Rightarrow n_{H_2}=n_S=n_{Fe}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l);m_S=0,2.32=6,4(g)$

Bài 11:

$PTHH:4Fe+3O_2\xrightarrow{t^o}2Fe_2O_3\\ Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ n_{Fe}=\dfrac{16,8}{56}=0,3(mol)\\ \Rightarrow n_{O_2}=\dfrac{3}{4}n_{Fe}=0,225(mol);n_{Fe_2O_3}=0,5n_{Fe}=0,15(mol)\\ \Rightarrow n_{H_2SO_4}=3n_{Fe_2O_3}=0,45(mol)\\ \Rightarrow V_{O_2}=0,225.22,4=5,04(l);m_{H_2SO_4}=0,45.98=44,1(g)$

Bài 12:

$PTHH:2H_2+O_2\xrightarrow{t^o}2H_2O\\ n_{H_2O}=\dfrac{900}{18}=50(mol)\\ \Rightarrow n_{H_2}=n_{H_2O}=50(mol);n_{O_2}=0,5.n_{H_2O}=25(mol)\\ \Rightarrow V_{H_2}=50.22,4=1120(l);V_{O_2}=25.22,4=560(l)$

Bài 13:

$PTHH:Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O\\ m_{Fe}=10.98\%=9,8(g)\Rightarrow n_{Fe}=\dfrac{9,8}{56}=0,175(mol)\\ \Rightarrow n_{Fe_2O_3}=0,0875(mol);n_{H_2}=0,2625(mol)\\ \Rightarrow m_{Fe_2O_3}=0,0875.160=14(g);V_{H_2}=0,2625.22,4=5,88(l)$

Bài 14:

$n_{Fe}=\dfrac{11,2}{56}=0,2(mol);n_{CuSO_4}=\dfrac{40}{160}=0,25(mol)\\ \text{Lập tỉ lệ: }\dfrac{0,2}{1}<\dfrac{0,25}{1}\Rightarrow CuSO_4\text{ dư}\\ \Rightarrow n_{Cu}=n_{Fe}=0,2(mol)\Rightarrow m_{Cu}=0,2.64=12,8(g)$