Giúp mk vs ạ. Mk đang cần gấp. Cảm ơn mn

15.

`(a)`

`200mL=0,2L`

`n_{NaOH}=0,2.1=0,2(mol)`

`NaOH+HCl->NaCl+H_2O`

Theo PT: `n_{NaCl}=n_{NaOH}=0,2(mol)`

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

`(b)`

`V_{dd\ NaCl}=0,2+0,2=0,4(L)`

`->C_{M\ NaCl}={0,4}/{0,2}=2M`

16.

Đặt 2 KL là `A:x(mol);B:y(mol)`

`A+2HCl->ACl_2+H_2`

`B+2HCl->BCl_2+H_2`

Do `C_{M\ ACl_2}=C_{M\ BCl_2},` cùng `V_{dd}`

`->n_{ACl_2}=n_{BCl_2}`

Theo PT: `n_A=n_{ACl_2};n_B=n_{BCl_2}`

`->n_A=n_B`

Mà `2n_A+2n_B=n_{HCl}=0,2.1,25=0,25`

`->n_A=n_B=0,0625(mol)`

`m_X=0,0625.M_A+0,0625.M_B=2,45`

`->M_A+M_B=39,2`

`->M_A,M_B<39,2`

Đáp án B thỏa `Be=9<39,2;Mg=24<39,2`