To determine the volume of CO gas formed from the complete decomposition of 444 grams of Ni(CO)₄ at a pressure of 752 torr and a temperature of [tex]\(22.0^\circ \text{C}\)[/tex], we can follow these steps:
### Step 1: Calculate the moles of Ni(CO)₄
First, we need the molar mass of Ni(CO)₄, which is 170.73 g/mol.
[tex]\[ \text{moles of Ni(CO)}_4 = \frac{\text{mass of Ni(CO)}_4}{\text{molar mass of Ni(CO)}_4} \][/tex]
[tex]\[ \text{moles of Ni(CO)}_4 = \frac{444 \text{ g}}{170.73 \text{ g/mol}} \][/tex]
This gives us:
[tex]\[ \text{moles of Ni(CO)}_4 = 2.6006 \text{ mol} \][/tex]
### Step 2: Calculate the moles of CO produced
From the balanced chemical equation [tex]\( \text{Ni(CO)}_4 \rightarrow \text{Ni} + 4 \text{CO} \)[/tex], we see that one mole of Ni(CO)₄ produces four moles of CO. Therefore:
[tex]\[ \text{moles of CO} = 4 \times \text{moles of Ni(CO)}_4 \][/tex]
[tex]\[ \text{moles of CO} = 4 \times 2.6006 \text{ mol} \][/tex]
This gives us:
[tex]\[ \text{moles of CO} = 10.4024 \text{ mol} \][/tex]
### Step 3: Use the Ideal Gas Law to find the volume of CO
The Ideal Gas Law is given by [tex]\( PV = nRT \)[/tex], where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm),
- [tex]\( V \)[/tex] is the volume in liters (L),
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the ideal gas constant, which is 0.0821 L⋅atm/(mol⋅K),
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
First, convert the temperature from Celsius to Kelvin:
[tex]\[ T = 22.0^\circ \text{C} + 273.15 = 295.15 \text{ K} \][/tex]
Next, convert the pressure from torr to atmospheres:
[tex]\[ P = \frac{752 \text{ torr}}{760 \text{ torr/atm}} = 0.9895 \text{ atm} \][/tex]
Now, solve for the volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
[tex]\[ V = \frac{10.4024 \text{ mol} \times 0.0821 \text{ L⋅atm/(mol⋅K)} \times 295.15 \text{ K}}{0.9895 \text{ atm}} \][/tex]
This gives us:
[tex]\[ V = 254.75 \text{ L} \][/tex]
Therefore, the volume of CO gas formed from the complete decomposition of 444 grams of Ni(CO)₄ at 752 torr and [tex]\(22.0^\circ \text{C}\)[/tex] is approximately 254.75 liters.
