Hỗn hợp khí A gồm CO,N2 và O2 (trong đó thể tích Ni tơ gấp 4 lần thể tích Oxi). Biết 33,6 lít hỗn hợp A ở đktc cân nặng 42,72g . Hãy tính % theo thể tích mỗi khí trong hỗn hợp A.

tranthithu20112005

2 năm trước

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phamhuu123

Gọi x là nCO, y là nO2, 4y là nN2

Do thể tích của nito gấp 4 lần thể tích của oxi, mà thể tích cũng giống như số mol nên nN2=4nO2=4y

Số mol hỗn hợp khí là:

`n=(33,6)/(22,4)=1,5mol`

Lập hệ PT:

`28x+32y+28.4y=42,72`

`x+y+4y=1,5`

Rút gọn ta được:

`28x+144y=42,72`

`x+5y=1,5`

`->x=0,6, y=0,18mol`

Vậy `n_(CO)=0,6, n_(O_2)=0,18, n_(N_2)=0,72mol`

`%V_(CO)=((0,6).100)/(1,5)=40%`

`%V_(O_2)=((0,18).100)/(1,5)=12%`

`->%V_(N_2)=100-40-12=48%`

Vote (0) Phản hồi (0) 2 năm trước

nguyentrung22602

Đặt $x$, $4y$, $y$ là số mol $CO$, $N_2$, $O_2$ trong $33,6l$ $A$

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

$\to x+4y+y=x+5y=1,5$ $(1)$

$m_A=42,72g$

$\to 28x+28.4y+32.y=42,72$ $(2)$

Từ $(1)(2)\to x=0,6; y=0,18$

$\to 1,5\rm mol A: \begin{cases} CO: 0,6(mol)\\ N_2: 0,72(mol)\\ O_2: 0,18(mol)\end{cases}$

$\%V_{CO}=\dfrac{0,6.100}{1,5}=40\%$

$\%V_{N_2}=\dfrac{0,72.100}{1,5}=48\%$

$\%V_{O_2}=12\%$

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