Đáp án + Giải thích các bước giải:

$a)PTHH:$

$2Ca+O_2\xrightarrow{t^o}2CaO$

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

$b)m_{Ca}=50.75\%=37,5(g)$

$m_{Mg}=50-37,5=12,5(g)$

$n_{Ca}=\dfrac{37,5}{40}=0,9375(mol)$

$n_{Mg}=\dfrac{12,5}{24}\approx 0,52(mol)$

Theo PT: $n_{CaO}=n_{Ca}=0,9375(mol);n_{MgO}=n_{Mg}=0,52(mol)$

$\Rightarrow \begin{cases} m_{CaO}=0,9375.56=52,5(g)\\ m_{MgO}=0,52.40=20,8(g)\end{cases}$

$c)\%m_{CaO}=\dfrac{52,5}{52,5+20,8}.100\%\approx 71,62\%$

$\%m_{MgO}=100-71,62=28,38\%$

$d)$ Theo PT: $n_{O_2}=0,5(n_{Mg}+n_{Ca})$

$\Rightarrow n_{O_2}=0,5(0,52+0,9375)=0,72875(mol)$

$\Rightarrow V_{O_2(đktc)}=0,72875.22,4=16,324(lít)$

Đáp án + Giải thích các bước:

$a/$

$2Ca+O_2 \xrightarrow{t^{o}} 2CaO$

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

$b/$

$n_{Ca}=\dfrac{50.75\%}{40}=0,9375(mol)$

$n_{Mg}=\dfrac{50.25\%}{24}=0,52$

$\to n_{CaO}=0,9375(mol)$

$n_{MgO}=0,52$

$m_{CaO}=52,5(g)$

$m_{MgO}=20,8(g)$

$c/$

$\%m_{CaO}=\dfrac{52,5}{52,20,8}.100\%=71,62\%$

$\%m_{MgO}=28,38\%$

$d/$

$n_{O_2}=0,5.0,9375+0,5.0,52=0,72875(mol)$

$V_{O_2}=0,72875.22,4=16,324(l)$