Đáp án + Giải thích các bước giải:

Bài 1:

 a) $6kg = 6000g$

$n_P =\dfrac{m}{M} = \dfrac{6000}{31} (mol)$

PTHH:
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
Theo PTHH:
$n_{O_2} =\dfrac{5}{4}n_P = \dfrac{5}{4} . \dfrac{6000}{31} = \dfrac{7500}{31}(mol)$
$\Rightarrow V_{O_2} =n . 22,4 = \dfrac{7500}{31} . 22,4 = \dfrac{168000}{31}(l)$

$\Rightarrow V_{kk} =V_{O_2} : \dfrac{1}{5} = \dfrac{168000}{31} : \dfrac{1}{5} = \dfrac{840000}{31}(l) \approx 27096,77(l)$

b) Theo PTHH:
$n_{P_2O_5} = \dfrac{1}{2}n_P = \dfrac{1}{2} . \dfrac{6000}{31} = \dfrac{3000}{31}(mol)$
$\Rightarrow m_{P_2O_5} = n . M = \dfrac{3000}{31} . (2 . 31 + 5 . 16) = \dfrac{3000}{31} . 142 = \dfrac{426000}{31} \approx 13741,94(g)$

Bài 2:

a) $n_{MgO} = \dfrac{m}{M} = \dfrac{4}{24 + 16} = \dfrac{4}{40} = 0,1(mol)$

PTHH:
$MgO + 2HCl \longrightarrow MgCl_2 + H_2O$

Theo PTHH:
$n_{HCl} = 2n_{MgO} = 2 . 0,1 = 0,2(mol)$
$\Rightarrow m_{HCl} = n . M = 0,2 . (35,5 + 1) = 0,2 . 36,5 = 7,3(g)$

$\Rightarrow C\%_{HCl} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{7,3}{73} . 100\% = 10\%$

Đáp án:

Bài 1:

$4P + 5O_{2}\xrightarrow{t^o}2P_{2}O_{5}$

a) Ta có: $6,2kg= 6200g$

⇒ $n_{P}=\dfrac{m}{M}=\dfrac{6200}{31}= 200 (mol)$

Theo PT: $n_{O_{2}}=\dfrac{5}{4}n_{P}=\dfrac{5}{4} × 200= 250 (mol)$

⇒ $V_{O_{2}}=n × 22,4=250 × 22,4= 5600 (l)$

⇒ $V_{\text{không khí}}=V_{O_{2}} × 5=5600 × 5= 28000 (l)$

b) Theo PT: $n_{P_{2}O_{5}}=\dfrac{2}{4}n_{P}=\dfrac{1}{2} × 200= 100 (mol)$

⇒ $m_{P_{2}O_{5}}=n × M=100 × 142= 14200 (g)$

Bài 2:

$MgO + HCl→MgCl_{2} + H_{2}O$

Ta có: $n_{MgO}=\dfrac{m}{M}=\dfrac{4}{40}= 0,1 (mol)$

Theo PT: $n_{HCl}=2n_{MgO}=2 × 0,1= 0,2 (mol)$

⇒ $m_{HCl}=n × M=0,2 × 36,5= 7,3 (g)$

⇒ $\text{C%HCl=$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{7,3}{73}$ × 100%= 10%}$