Câu 9:

$n_{HCl}=\dfrac{146.5\%}{100\%.36,5}=0,2(mol)\\ PTHH:CaCO_3+2HCl\to CaCl_2+H_2O+CO_2\uparrow\\ a,n_{CaCO_3}=\dfrac{1}{2}n_{HCl}=0,1(mol)\\ \Rightarrow m_{CaCO_3}=0,1.100=10(g)\\ b,n_{CO_2}=n_{CaCO_3}=0,1(mol)\\ \Rightarrow V_{CO_2}=0,1.22,4=2,24(l)\\ c,PTHH:Ca(OH)_2+2HCl\to CaCl_2+H_2O\\ \Rightarrow n_{Ca(OH)_2}=n_{CaCl_2}=\dfrac{1}{2}n_{HCl}=0,1(mol)\\ \Rightarrow m_{Ca(OH)_2}=0,1.74=7,4(g);m_{CaCl_2}=0,1.111=11,1(g)\\ \Rightarrow C\%_{CaCl_2}=\dfrac{11,1}{7,4+146}.100\%=7,236\%$

Câu 10:

$a,PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ b,n_{Zn}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Zn}=0,1.65=6,5(g)\\ \Rightarrow m_{\text{rắn sau p/ứ}}=m_{Cu}=10,5-6,5=4(g)\\ c,n_{Cu}=\dfrac{4}{64}=0,0625(mol)\\ PTHH:Zn+2AgNO_3\to Zn(NO_3)_2+2Ag\\ Cu+2AgNO_3\to Cu(NO_3)_2+2Ag\\ \Rightarrow \Sigma n_{Ag}=2n_{Zn}+2n_{Cu}=0,2+0,125=0,325(mol)\\ \Rightarrow m_{Ag}=0,325.108=35,1(g)$