a)

n_{CO_2}={4,48}/{22,4}=0,2(mol)

CaO+2HCl->CaCl_2+H_2O

CaCO_3+2HCl->CaCl_2+CO_2+H_2O

Theo PT: n_{CaCO_3}=n_{CO_2}=0,2(mol)

->\%m_{CaCO_3}={0,2.100}/{36,8}.100\%\approx 54,35\%

->\%m_{CaO}=100-54,35=45,65\%

b)

n_{CaO}={36,8-0,2.100}/{56}=0,3(mol)

Theo PT: n_{CaCl_2}=n_{CaO}+n_{CaCO_3}=0,5(mol)

->C_{M\ CaCl_2}={0,5}/5=0,1M

c)

2NaCl+H_2SO_4  $$\xrightarrow{t^o}$$  Na_2SO_4+2HCl

2KCl+H_2SO_4  $$\xrightarrow{t^o}$$  K_2SO_4+2HCl

2HCl+Zn->ZnCl_2+H_2

Đặt n_{NaCl}=x(mol);n_{KCl}=y(mol)

->58,5x+74,5y=26,6(1)

n_{H_2}={4,48}/{22,4}=0,2(mol)

Theo PT: x+y=n_{HCl}=2n_{H_2}=0,4(2)

(1)(2)->x=y=0,2(mol)

m_{NaCl}=0,2.58,5=11,7(g)

->m_{KCl}=26,6-11,7=14,9(g)

\%m_{NaCl}={11,7}/{26,6}.100\%\approx 43,98\%

->\%m_{KCl}=100-43,98=56,02\%

a)

n_{CO_2} = (4,48)/(22,4) = 0,2(mol)

CaO + 2HCl -> CaCl_2 + H_2O

CaCO_3 + 2HCl -> CaCl_2 +CO_2 + H_2O

Theo PT: n_{CaCO_3} = n_{CO_2} = 0,2(mol)

=> m_{CaCO_3} = 0,2.100 = 20(g)

=> m_{CaO} = 36,8 - 20 = 16,8(g)

\%m_{CaCO_3} = (20)/(36,8) .100\% = 54,35\%

\%m_{CaO} = 100\% - 54,35\% = 45,65\%

b)

n_{CaO} = (16,8)/(56) = 0,2(mol)

Theo PT: n_{CaCl_2} = n_{CaCO_3} + n_{CaO} = 0,4(mol)

=> C_{M(CaCl_2)} = (0,4)/5 = 0,08M

c)

2NaCl + H_2SO_4 -> Na_2SO_4 + 2HCl

2KCl + H_2SO_4 -> K_2SO_4 + 2HCl

NaCl + H_2SO_4 -> NaHSO_4 + HCl

KCl + H_2SO_4 -> KHSO_4 + HCl

Zn + 2HCl-> ZnCl_2 + H_2

Gọi n_{NaCl} = a(mol); n_{KCl} = b(mol)

=> 58,5a + 74,5b = 26,6(1)

n_{H_2} = (4,48)/(22,4) = 0,2(mol)

Theo PT: n_{HCl} = 2n_{H_2} = 0,4(mol0

n_{HCl} = n_{NaCl} + n_{KCl}

=> a + b = 0,4(2)

(1),(2) => a = b = 0,2

m_{NaCl} = 0,2.58,5 = 11,7(g)

m_{KCl} = 0,2.74,5 = 14,9(g)

\%m_{NaCl} = (11,7)/(11,7 + 14,9) .100\% = 44\%

\%m_{KCl} = 100\% - 44\% = 56\%