Bài 1.
PTHH: $Mg+2HCl→MgCl_2+H_2$
a/ $n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{3,6}{24}=0,15(mol)$
Theo pt: $→n_{H_2}=0,15(mol)$
$→V_{H_2}=n_{H_2}.22,4=3,36(l)$
b/ Theo pt: $→n_{HCl}=0,3(mol)$
$→m_{HCl}=n_{HCl}.M_{HCl}=0,3.36,5=10,95(g)$
c/ Theo pt: $→n_{MgCl_2}=0,15(mol)$
$→m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,15.95=14,25(g)$
Bài 2.
PTHH: $4P+5O_2→2P_2O_5$
$n_P=\dfrac{m_P}{M_P}=\dfrac{1,24}{32}=0,04(mol)$
Theo pt: $→n_{P_2O_5}=0,02(mol)$
$→m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,02.144=2,88(g)$
Theo pt: $→n_{O_2}=0,05(mol)$
$→V_{O_2}=n_{O_2}.22,4=1,12(l)$
