Answer:
0.908 L.
Explanation:
- Firstly, we should calculate the number of moles of 1.60 g argon.
- n of Ar = mass / molar mass of Ar = (1.60 g) / (39.948 g/mol) = 0.04 mol.
- |t is established and known that the molar volume of a gas at STP (Standard Temperature and Pressure) is equal to 22.4 L for 1 mole of any ideal gas at a temperature of 273.15 K and a pressure of 1.00 atm.
<u><em>using cross multiplication:</em></u>
1.0 mole of Ar gas will occupy → 22.7 L
0.04 mole of Ar gas will occupy → ??? L
∴ The volume occupied by Ar = (0.04 mole) (22.7 L) / (1.0 mole) = 0.908 L.
Answer:
Explanation:
Data:
Actual yield = 16.8 g
Theoretical yield = 49.7 g
Calculation:
I think d not sure tho but hope it helps
d= 8/10 so yah that was a guess
2C₃H₇OH + 9O₂ = 6CO₂ + 8H₂O
V(O₂)=12.0 dm³
n(C₃H₇OH)=0.1 mol
n(O₂)=12.0 dm³/22.4 dm³/mol=0.5357 mol
C₃H₇OH : O₂ 2:9 1:4.5
0.1:0.5357
oxygen in excess
V(CO₂)=3Vm*n(C₃H₇OH)
V(CO₂)=3*22.4*0.1=6.72 dm³
