Giải thích các bước giải:
Bài 3:
a. PTHH: $2Cu + O_2 \overset{t^{\circ}}{\rightarrow} 2CuO$
b. $n_{CuO}=$ $\dfrac{48}{80}=0,6(mol)$
Theo PTHH $\rightarrow n_{Cu}=n_{CuO}=0,6(mol)$
$\rightarrow m_{Cu}=0,6*64=38,4(g)$
$\rightarrow$ $H=\dfrac{38,4}{51,2}*100$% = 75%
Bài 4:
PTHH: $2SO_2 + O_2 \xrightarrow[t^{\circ}]{V_2O_5} 2SO_3$
$n_{SO_2}=$ $\dfrac{11,2}{22,4}=0,5(mol)$
Vì $H=80$% $\rightarrow$ $n_{SO_2pu}=0,5*80$% $=0,4(mol)$
Theo PTHH $\rightarrow$ $n_{SO_2}=n_{SO_3}=0,4(mol)$
$\rightarrow$ $m_{SO_3}=0,4*80=32(g)$