2
C
4
H
10
(
g
) + 13
O
2
(
g
) = 8
C
O
2
(
g
) + 10
H
2
O
(
g
)
Answer:
1120 gm
Explanation:
6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
(a) Balance the equation.
(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete
combustion reaction?
FIRST, CORRECT THE EQUATION THEN BALANCE
2C2H6(G) + 7O2------------> 4CO2 + 6H2O
so for 10 moles of ethane, we need
7 X 5 = 35 MOLES O2
=35 MOLES O2
O2 HAS A MOLAR MASS OF 2X16 = 32 gm
35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm
Answer:
0.007 mol
Explanation:
We can solve this problem using the ideal gas law:
PV = nRT
where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.
Keep in mind that when we collect a gas over water we have to correct for the vapor pressure of water at the temperature in the experiment.
Ptotal = PH₂O + PO₂ ⇒ PO₂ = Ptotal - PH₂O
Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.
P H₂O = 23.8 mmHg x 1 atm/760 mmHg = 0.031 atm
V = 1750 mL x 1 L/ 1000 mL = 0.175 L
T = (25 + 273) K = 298 K
PO₂ = 1 atm - 0.031 atm = 0.969 atm
n = PV/RT = 0.969 atm x 0.1750 L / (0.08205 Latm/Kmol x 298 K)
n = 0.007 mol
