`n_(MgO)=10/40=0,25(mol)`

$a)PTHH :$

$2Mg+O_2\xrightarrow{t^0}2MgO$

$b)$

Theo `PT,n_(Mg)=n_(MgO)=0,25(mol)`

`=>m_(Mg)=0,25.24=6(g)`

$c)$

Theo `PT,n_(O_2)=1/2n_(MgO)=0,125(mol)`

`=>V_(O_2)(đktc)=0,125.22,4=2,8(l)`

`=>V_(kk)=2,8.5=14(l)`

$d)$

$PT:Mg+2HCl \to MgCl_2+H_2$

Theo `PT,n_(HCl)=2n_(Mg)=0,5(mol)`

`=>m_(ctHCl)=0,5.36,5=18,25(g)`

`m_(ddHCl)=(18,25.100)/10=182,5(g)`

Đáp án + Giải thích các bước giải:

$a\big)2Mg+O_2\xrightarrow{t^o}2MgO$

$b\big)$

$n_{MgO}=\frac{10}{40}=0,25(mol)$

Theo PT: $n_{Mg}=n_{MgO}=0,25(mol)$

$\to m_{Mg}=0,25.24=6(g)$

$c\big)$

Theo PT: $n_{O_2}=\frac{1}{2}n_{Mg}=0,125(mol)$

$\to V_{kk}=0,125.5.22,4=14(l)$

$d\big)$

$Mg+2HCl\to MgCl_2+H_2$

Theo PT: $n_{HCl}=2n_{Mg}=0,5(mol)$

$\to m_{dd\,HCl}=\frac{0,5.36,5}{10\%}=182,5(g)$