Đáp án:

Giải thích các bước giải:

a)nO2=7,8422,4=0,35(mol)a)nO2=7,8422,4=0,35(mol)

Đặt nAl=x(mol);nZn=y(mol)nAl=x(mol);nZn=y(mol)

⇒27x+65y=mhh=31,4(1)⇒27x+65y=mhh=31,4(1)

PTHH:2Zn+O2to→2ZnOPTHH:2Zn+O2→to2ZnO

4Al+3O2to→2Al2O34Al+3O2→to2Al2O3

Theo PT: 0,75x+0,5y=nO2=0,35(2)0,75x+0,5y=nO2=0,35(2)

Từ (1)(2)⇒x=0,2(mol);y=0,4(mol)(1)(2)⇒x=0,2(mol);y=0,4(mol)

⇒mAl=0,2.27=5,4(g)⇒mAl=0,2.27=5,4(g)

⇒mZn=31,4−5,4=26(g)⇒mZn=31,4−5,4=26(g)

b)b) Theo 2 PT: nAl2O3=12x=0,1(mol)nAl2O3=12x=0,1(mol)

nZnO=y=0,4(mol)nZnO=y=0,4(mol)

⇒mAl2O3=0,1.102=10,2(g)⇒mAl2O3=0,1.102=10,2(g)

mZnO=0,4.81=32,4(g)

XIN CTLHN NHA

$a)n_{O_2}=\dfrac{7,84}{22,4}=0,35(mol)$

Đặt $n_{Al}=x(mol);n_{Zn}=y(mol)$

$\Rightarrow 27x+65y=m_{hh}=31,4(1)$

$PTHH:2Zn+O_2\xrightarrow{t^o}2ZnO$

$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

Theo PT: $0,75x+0,5y=n_{O_2}=0,35(2)$

Từ $(1)(2)\Rightarrow x=0,2(mol);y=0,4(mol)$

$\Rightarrow m_{Al}=0,2.27=5,4(g)$

$\Rightarrow m_{Zn}=31,4-5,4=26(g)$

$b)$ Theo 2 PT: $n_{Al_2O_3}=\dfrac{1}{2}x=0,1(mol)$

$n_{ZnO}=y=0,4(mol)$

$\Rightarrow m_{Al_2O_3}=0,1.102=10,2(g)$

$m_{ZnO}=0,4.81=32,4(g)$