`1.`

`a)`

`n_{K_2O}={2,35}/{94}=0,025(mol)`

`K_2O+H_2O->2KOH`

Theo PT: `n_{KOH}=2n_{K_2O}=0,05(mol)`

`->C\%_{KOH}={0,05.56}/{2,35+100}.100\%≈2,74\%`

`V_{dd\ KOH}={100+2,35}/{1,08}≈95(ml)=0,095(l)`

`->C_{M\ KOH}={0,05}/{0,095}≈0,53M`

`b)`

`2KOH+H_2SO_4->K_2SO_4+2H_2O`

Theo PT: `n_{H_2SO_4}=1/2n_{KOH}=0,025(mol)`

`->m_{dd\ H_2SO_4}={0,025.98}/{20\%}=12,25(g)`

`->V_{dd\ H_2SO_4}={12,25}/{1,15}≈10,65(ml)`

`2.`

`a)`

Đặt `n_{Al}=2x(mol)->n_{Mg}=x(mol)`

`->27.2x+24x=7,8`

`->x=0,1(mol)`

`->n_{Al}=0,2(mol);n_{Mg}=0,1(mol)`

`b)`

`m_{Mg}=0,1.24=2,4(g)`

`m_{Al}=7,8-2,4=5,4(g)`

Đáp án+ Giải thích các bước giải:

$\rm 1a)$
$\rm K_2O+H_2O\to 2KOH$
$\rm n_{K_2O}=\dfrac{2,35}{94}=0,025(mol)$
$\rm n_{KOH}=2n_{K_2O}=0,05(mol)$
$\rm V_{dd\ KOH}$`={2,35+100}/{1,08}=`$\rm 94,76(ml)=0,09476(l)$
$\rm C_{M\ dd\ KOH}=\dfrac{0,05}{0,09476}$`~~`$\rm 0,53M$
$\rm C$`%`$\rm _{dd\ KOH}$`={0,05.56}/{2,35+100}.100%~~2,73%`
$\rm 1b)$
$\rm 2KOH+H_2SO_4\to K_2SO_4+2H_2O$
$\rm n_{H_2SO_4}=0,5n_{KOH}=0,025(mol)$
$\rm m_{dd\ H_2SO_4}=\dfrac{0,025.98}{20%}=12,25(g)$
$\rm V_{dd\ H_2SO_4}=\dfrac{12,25}{1,15}$`~~`$\rm 10,65(ml)$

$\rm 2)$
$\rm  n_{Mg}:a(mol)\ , n_{Al}:2a(mol)$
$\rm m_Y=24a+27.2a=7,8$
`=>`$\rm a=0,1(mol)$
$\rm n_{Al}=2.0,1=0,2(mol)\ , n_{Mg}=0,1(mol)$
$\rm m_{Al}=0,2.27=5,4(g)$
$\rm m_{Mg}=7,8-5,4=2,4(g)$