If a 28.0 L balloon with a temperature of
1 answer:
You might be interested in
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 %
<em>What volume do 5 moles of a gas occupy at 28 ° C and 3 atm of pressure?</em>
<em />
<h3>Further explanation</h3>In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
n= 5 moles
T=28 +273=301 K
P=3 atm
The volume of the gas :