Đáp án + Giải thích các bước giải:

Bài 5:

$a\big)Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O$

$b\big)$

$n_{Fe_3O_4}=\frac{23,2}{232}=0,1(mol)$

Theo PT: $n_{Fe}=3n_{Fe_3O_4}=0,3(mol)$

$\to m_{Fe}=0,3\times 56=16,8(g)$

Theo PT: $n_{H_2}=4n_{Fe_3O_4}=0,4(mol)$

$\to V_{H_2(đkc)}=0,4\times 24,79=9,916(l)$

$c\big)$

$2Al+6HCl\to 2AlCl_3+3H_2$

Theo PT: $n_{Al}=\frac{2}{3}n_{H_2}=\frac{4}{15}(mol)$

$\to m_{Al}=\frac{4}{15}\times 27=7,2(g)$

Bài 6:

$a\big)2Ca+O_2\xrightarrow{t^o}2CaO$

$b\big)$

$n_{CaO}=\frac{42}{56}=0,75(mol)$

Theo PT: $n_{Ca}=n_{CaO}=0,75(mol)$

$\to m=m_{Ca}=0,75\times 40=30(g)$

$c\big)$

Theo PT: $n_{O_2}=\frac{1}{2}n_{Ca}=0,375(mol)$

$\to V_{O_2(đkc)}=0,375\times 24,79=9,29625(l)$

$\to V_{\rm không\,khí(đkc)}=9,29625\times 5=46,48125(l)$

Đáp án + Giải thích các bước giải:

 a) PTHH:
$Fe_3O_4 + 4H_2 \overset{t^o}{\longrightarrow} 3Fe + 4H_2O(1)$
b) $n_{Fe_3O_4} = \dfrac{m}{M} = \dfrac{23,2}{3 . 56 + 4 . 16} = \dfrac{23,2}{232} = 0,1\text{ mol}$

Theo PTHH(1):
$n_{Fe} = 3n_{Fe_3O_4} = 3 . 0,1 = 0,3\text{ mol}$
$\Rightarrow m_{Fe} = n . M = 0,3 . 56 = 16,8g$

$n_{H_2} = 4n_{Fe_3O_4} = 4 . 0,1 = 0,4\text{ mol}$

$\Rightarrow V_{H_2} = n . 24,79 = 0,4 . 24,79 = 9,916l$

c) PTHH:

$2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2 \uparrow(2)$

Theo PTHH(2):

$n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{2}{3} . 0,4 = \dfrac{0,8}{3}\text{ mol}$

$m_{Al} = n . M = \dfrac{0,8}{3} . 27 = 7,2g$

Bài 6:
a) PTHH:
$2Ca + O_2 \overset{t^o}{\longrightarrow} 2CaO$

b) $n_{CaO} = \dfrac{m}{M} = \dfrac{42}{40 + 16} = \dfrac{42}{56} = 0,75\text{ mol}$
Theo PTHH:
$n_{Ca} = n_{CaO} = 0,75\text{ mol}$

$\Rightarrow m = m_{Ca} = n . M = 0,75 . 40 = 30g$

c) Theo PTHH:
$n_{O_2} = \dfrac{1}{2}n_{Ca} = \dfrac{1}{2} . 0,75 = 0,375\text{ mol}$

$\Rightarrow V_{O_2} = n . 24,79 = 0,375 . 24,79 = 9,29625l$

$V_{kk} = 5V_{O_2} = 5 . 9,29625 = 46,48125l$