Đáp án:
Câu 1: \[C_xH_y+\bigg(x+\dfrac{y}{4}\bigg)O_2\to xCO_2+\dfrac{y}{2}H_2O\]
Câu 2:
$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$
a. PTHH: $Fe+2HCl\to FeCl_2+H_2$
b. Theo PTHH: $n_{H_2}=n_{Fe}=0,2(mol)$
$\to V_{H_2}=0,2\times 22,4=4,48(l)$
c. Theo PTHH: $n_{HCl}=2\times n_{Fe}=0,2\times 2=0,4(mol)$
$\to m_{HCl}=0,4\times 36,5=14,6(g)$
d. Theo PTHH: $n_{Fe}=n_{FeCl_2}=0,2(mol)$
$\to m_{FeCl_2}=0,2\times 127=25,4(g)$
Câu 3:
$n_{Mg}=\dfrac{6}{24}=0,25(mol)$
$n_{O_2}=\dfrac{2,24}{22,4}=0,1(mol)$
a. PTHH: $2Mg+O_2\to 2MgO$
Vì: $\dfrac{0,25}{2}>\dfrac{0,1}{1}\to$ Sau phản ứng $O_2$ hết, $Mg$ dư
b. Theo PTHH: $n_{MgO}=2\times n_{O_2}=2\times 0,1=0,2(mol)$
$\to m_{MgO}=40\times 0,2=8(g)$