$n_{NaCl}=\dfrac{5.85}{23+35.5}=0.1(mol)$

Gọi $x$ là số mol $Na_2O.$

$Na_2O+2HCl \to 2NaCl+H_2O$

 $x$  `->` $2x$  `->` $2x$

`->`$n_{NaCl}=2x(mol);n_{HCl,pu}=2x(mol)$

`=>`$2x=0.1$

`=>`$x=0.05$

`=>`$m_{Na_2O}=nM=0.05 \times 62=3.1(gam)$

$n_{HCl,dư}=V \times C_M=0.1 \times 0.01=0.001(mol)$

`->`$n_{HCl}=0.001+2 \times 0.05=0.101(mol)$

`=>`$C_M(HCl)=\dfrac{0.101}{0.1}=1.01(M)$

`n_{NaCl} = (5,85)/(58,5) = 0,1(mol)`

`Na_2O + 2HCl -> 2NaCl + H_2O`

`n_{HCl(dư)} = 0,01.0,1 = 0,001(mol)`

Theo PT: `n_{HCl(pư)} = n_{NaCl} = 0,1 (mol)`

`n_{HCl(bđ)} = 0,001 + 0,1 = 0,101(mol)`

`=> C_{M(HCl)} = (0,101)/(0,1) = 1,01M`