a)

Đặt n_{NaOH}=x(mol);n_{Ca(OH)_2}=y(mol)

->40x+74y=7,7(1)

200mL=0,2L->n_{HCl}=0,2.1=0,2(mol)

HCl+NaOH->NaCl+H_2O

2HCl+Ca(OH)_2->CaCl_2+2H_2O

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

(1)(2)->x=0,1(mol);y=0,05(mol)

->m_{NaOH}=0,1.40=4(g)

->m_{Ca(OH)_2}=7,7-4=3,7(g)

b)

Theo PT:

n_{NaCl}=x=0,1(mol)

n_{CaCl_2}=y=0,05(mol)

->m_{NaCl}=0,1.58,5=5,85(g)

        m_{CaCl_2}=0,05.111=5,55(g)

c)

C_{M\ NaCl}={0,1}/{0,2}=0,5M

C_{M\ CaCl_2}={0,05}/{0,2}=0,25M