Question

Answer 1

Câu 14 : 

a)

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
$2H_2 + O_2 \xrightarrow{t^o} 2H_2O$

b)

Theo PTHH : 

$V_{CH_4} = V_{CO_2} = 5,04(lít)$

$\%V_{CH_4} = \dfrac{5,04}{8,96}.100\% = 56,25\%$
$\%V_{H_2} = 100\% -56,25\% = 43,75\%$

c)

$V_{H_2} = 8,96.43,75\% = 3,92(lít)$

$V_{O_2} = 2V_{CH_4} + 0,5V_{H_2} = 12,04(lít)$
$n_{O_2} = \dfrac{12,04}{22,4} = 0,5375(mol)$

$2H_2O \xrightarrow{đp}2H_2  + O_2$

$n_{H_2O} = 2n_{O_2} = 1,075(mol)$

Số phân tử H2O là 1,075.6.10²³ = 6,45.10²³

Answer 2

Câu 15 : 

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

$n_C = n_{O_2} = \dfrac{8680.20\%}{22,4} = 77,5(mol)$

$m_C = 77,5.12 = 930(gam) = 0,93(kg)$

$\%m_{tạp\ chất} = \dfrac{1-0,93}{1}.100\% = 0,07\%

Answer 3

Câu 16 : 

$n_{Fe} = \dfrac{2,24}{56} = 0,04(mol)$

$n_{O_2} = \dfrac{2,688.20\%}{22,4} = 0,024(mol)$

$3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$

Ta thấy :

$n_{Fe} : 3 = 0,013 > n_{O_2} : 2 = 0,012$ nên Fe dư

$n_{Fe\ pư} = \dfrac{3}{2}n_{O_2} = 0,036(mol)$

$n_{Fe_3O_4} = \dfrac{1}{2}n_{O_2} = 0,012(mol)$

Sau phản ứng :

$m_{Fe_3O_4} = 0,012.232 = 2,784(gam)$

$m_{Fe\ dư} = (0,04 - 0,036).56 = 0,224(gam)$