Respuesta :
Answer:
[tex]V=52.8L[/tex]
Explanation:
Hello,
In this case, given the reaction:
[tex]2PbS(s) + 3O_2(g) \rightarrow 2PbO(s) + 2SO_2(g)[/tex]
827.4 kJ of heat are releasing 2 moles of sulfur dioxide, thus, for 975 kJ we find the following released moles:
[tex]n=\frac{-975 kJ*2mol}{-827.4kJ} =2.36mol[/tex]
Then, at STP conditions (273K and 1atm) we compure the volume by using the ideal gas equation:
[tex]V=\frac{nRT}{P}=\frac{2.36mol*0.082\frac{atm*L}{mol*K}*273K}{1atm} \\\\V=52.8L[/tex]
Regards.
The volume of sulphur dioxide, SO₂ produced at STP if 975 KJ of heat are liberated is 52.9 L
We'll begin by calculating the number of mole of SO₂ produced when 975 KJ of heat are liberated. This can be obtained as follow:
2PbS + 3O₂ —> 2PbO + 2SO₂ ∆H = –827.4 KJ
From the balanced equation above,
827.4 KJ of heat energy produced 2 moles of SO₂.
Therefore,
975 KJ of heat will produce = (975 × 2)/ 827.4 = 2.36 moles of SO₂.
- Finally, we shall determine the volume of SO₂ produced from the reaction. This can be obtained as follow:
Number of mole (n) = 2.36 moles
Pressure (P) = STP = 1 atm
Temperature (T) = STP = 273 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
1 × V = 2.36 × 0.0821 × 273
V = 52.9 L
Thus, the volume of sulphur dioxide, SO₂ produced at STP if 975 KJ of heat are liberated is 52.9 L
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