The volume of CO2 prodiced when 12 mL of CH4 and 8 mL of C4H10 is burnt is 56 mL.
What is the equation of the reaction?
The equation of the combustion of C2H6, CH4, C4H10, and C3H8 gases are given below:
2 C2H6 + 7 O2 ---> 4 CO2 + 6 H20
C3H8 + 5 O2 ---> 3 CO2 + 4 H20
CH4 + 2 O2 ----> CO2 + 2 H2O
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O
X moles of C2H6 will produce 2X moles of CO2
Y moles of C3H8 will produce 3Y moles of CO2
Thus;
X + Y = 20 mL
X =20 - Y ---(1)
2X + 3Y = 52 mL ---(2)
Solving for X by substituting in (2)
2 (20 - Y) + 3Y = 52
40 - 2Y + 3Y = 52
Y = 12 mL
Then X = 8 mL
Volume of CO2 produced when 12 mL of CH4 and 8 mL of C4H10 is burnt = (12 × 2 + 8 × 4) = 56 mL
Therefore, the volume of CO2 produced is 56 mL.
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