1.0 mole of gas at STP occupies 22.4 Litres.
4/0L / 22.4 L/mole = 0.179 moles He.
Answer:
The answer to your question is: 70.8%
Explanation:
Data
Al₂O₃ = 60 g
C = 30 g
CO = gas
Al = 22.5 g
MW Al₂O₃ = 102 g
MW C = 12 g
MW Al = 54 g
Reaction
Al₂O₃ + 3C ⇒ 3 CO + 2 Al
Limiting reactant
102 g of Al₂O₃ -------------- 54 g Al
60 g -------------- x
x = 31.8 g
36 g of C ------------------ 54 g of Al
30 g of C ------------------ x
x = 45 g of Al
Limiting reactant = Al₂O₃
Percent yield = 
Percent yield = 70.75 %
Answer: PV = nRT
A gas at STP... This means that the temperature is 0°C and pressure is 1 atm.
R is the gas constant which is 0.08206 L*atm/(K*mol)
Rearranging for volume
V = nRT/P
The temperature and number of moles are held constant. This means that this uses Boyle's Law. (The ideal gas law could be manipulated to give us this result when T and n are held constant.)
PV = k
where k is a constant.
This means that
P₁V₁ = k = P₂V₂
P₁V₁ = P₂V₂
(1 atm) * (1 L) = (2 atm) * V₂
V₂ = 0.5 L
The new volume of the gas is 0.5 L.
Explanation:
The answer is C. Assume specific heat to be 4.18 J/g/C