Đáp án:

a) $m_{O_2}=60(g)$

b) $m_{O_2}=80(g)$

c) $m_{O_2}=60(g)$

d) $m_{O_2}=24(g)$

Giải thích các bước giải:

 a)

$4P +5O_2 \xrightarrow{t^o} 2P_2O_5$

$n_{P}=\dfrac{46,5}{31}=1,5(mol)$

Theo ptpư: $n_{O_2}=\dfrac{5}{4}n_{P}=\dfrac{5}{4}.1,5=1,875(mol)$

$\to m_{O_2}=1,875.32=60 (g)$

b)

$C + O_2 \xrightarrow{t^o} CO_2$

$n_{C}=\dfrac{30}{12}=2,5(mol)$

Theo ptpư: $n_{O_2}=n_C=2,5(mol)$

$\to m_{O_2}=2,5.32=80 (g)$

c)

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

$n_{Al}=\dfrac{67,5}{27}=2,5 (mol)$

Theo ptpư: $n_{O_2}=\dfrac{3}{4}n_{Al}=1,875(mol)$

$\to m_{O_2}=1,875.32=60 (g)$

d)

$2H_2 + O_2 \xrightarrow{t^o} 2H_2O$

$n_{H_2}=\dfrac{33,6}{22,4}=1,5 (mol)$

Theo ptpư: $n_{O_2}=\dfrac{1}{2}n_{H_2}=0,75 (mol)$

$\to m_{O_2}=0,75.32=24(g)$

`a)` `n_{P}=\frac{46,5}{31}=1,5(mol)`

PTHH: `4P+5O_2\overset{t^o}{\to}2P_2O_5`

Theo phương trình, ta thấy: `n_{O_2}=n_P.\frac{5}{4}=1,5.1,25=1,875(mol)`

`\to m_{O_2}=1,875.32=60g`

 `b)` `n_{C}=\frac{30}{12}=2,5(mol)`

PTHH: `C+O_2\overset{t^o}{\to}CO_2`

Theo phương trình, ta thấy: `n_C=n_{O_2}=2,5(mol)`

`\to m_{O_2}=2,5.32=80g`

`c)` `n_{Al}=\frac{67,5}{27}=2,5(mol)`

PTHH: `4Al+3O_2\overset{t^o}{\to}2Al_2O_3`

Theo phương trình, ta thấy: `n_{O_2}=0,75.n_{Al}=0,75.2,5=1,875(mol)`

`\to m_{O_2}=1,875.32=60g`

`d)` `n_{H_2}=\frac{33,6}{22,4}=1,5(mol)`

PTHH: `2H_2+O_2\overset{t^o}2H_2O`

Theo phương trình, ta thấy: `n_{O_2}=\frac{n_{H_2}}{2}=\frac{1,5}{2}=0,75(mol)`

`\to m_{O_2}=0,75.32=24g`