Câu 3:

$n_{CH_4}=\dfrac{1,12}{22,4}=0,05(mol)$

$PTHH:CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

Theo PT: $n_{CO_2}=n_{CH_4}=0,05(mol);n_{O_2}=2n_{CH_4}=0,1(mol)$

$\Rightarrow V_{CO_2(đktc)}=0,05.22,4=1,12(l)$

$\Rightarrow V_{O_2(đktc)}=0,1.22,4=2,24(l)$

Câu 4:

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2\uparrow$

$b)n_{Al}=\dfrac{6,75}{27}=0,25(mol)$

Theo PT: $n_{H_2}=\dfrac{3}{2}n_{Al}=0,375(mol)$

$\Rightarrow V_{H_2(đktc)}=0,375.22,4=8,4(l)$

$c)$ Theo PT: $n_{HCl}=3n_{Al}=0,75(mol)$

$\Rightarrow m_{HCl}=0,75.36,5=27,375(g)$

$d)$ Theo PT: $n_{Al}=n_{AlCl_3}=0,25(mol)$

$\Rightarrow m_{AlCl_3}=0,25.133,5=33,375(g)$

Đáp án+Giải thích các bước giải:

 Câu 4:

PTHH: CH4+2O2---->CO2+2H2O

Ta có: nCH4=1,12÷22,4= 0,05 mol

Theo PTHH ta có: nO2=2×0,05= 0,1 mol

⇒VO2=0,1×22,4= 2,24 lít

Theo PTHH ta có: nCO2=nCH4= 0,05 mol

⇒VCH4=0,05×22,4= 1,12 lít

Câu 5:

a) PTHH: 2Al+6HCl----->2AlCl3+3H2

b) Ta có: nAl=6,75÷27= 0,25 mol

Theo PTHH ta có: nH2=3÷2×0,25= 0,375 mol

⇒VH2=0,375×22,4= 8,4 lít

c) Theo PTHH ta có: nHCl=6÷2×0,25= 0,75 mol

⇒mHCl=0,75×36,5= 27,375 g

d) Theo PTHH ta có: nAlCl3=nAl= 0,25 mol

⇒mAlCl3=0,25×133,5= 33,375 g