The volume of oxygen required to burn 12.00 L ethane is calculated as follows
find the moles of C2H6 used
At STP 1 mole is always = 22.4 L, what about 12.00 L
= ( 12.00L x 1 moles) 22.4 L = 0.536 moles
write the reacting equation
2C2H6+ 7O2 = 4CO2 + 6H2O
by use of mole ratio between C2H6 :O2 which is 2:7 the moles of O2
= 0.536 x7/2= 1.876 moles
again at STP 1mole = 22.4 L what about 1.876 moles
= 22.4 L x 1.876 moles/ 1 mole = 42.02 L
Where is the table? I dont know what to classify
Answer:
The answer to your question is: V2 = 1.94 l
Explanation:
Data
V1 = 2.42 l
T1 = 25°C
P1 = 1 atm
V2 = ?
T2 = 25 -11 = 14°C
P2 = 1(0.7) = 0.7 atm
Formula
P1V1/T1 = P2V2/T2
Clear V2 from the equation
V2 = P1V1T2/ P2T1
V2 = (1)((2.42)(14) / (0.7)(25)
V2 = 33.88 / 17.5
V2 = 1.94 l
Answer:
Don’t change, keep the same
Explanation:
Answer:
0.6 grams of hydrogen are needed to react with 2.75 g of nitrogen.
Explanation:
When hydrogen and nitrogen react they form ammonia.
Chemical equation:
N₂ + 3H₂ → 2NH₃
Given mass of nitrogen = 2.75 g
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2.75 g / 28 g/mol
Number of moles = 0.098 mol
Now we will compare the moles of nitrogen with hydrogen from balance chemical equation:
N₂ : H₂
1 : 3
0.098 : 3×0.098 = 0.3 mol
Mass of hydrogen:
Mass = number of moles × molar mass
Mass = 0.3 mol × 2 g/mol
Mass = 0.6 g