A mixture of C3H8 and C2H2 has a mass of 2.0 . It is burned in excess O2 to form a mixture of water and carbon dioxide that contains 1.5 times as many moles of CO2 as of water. Find the mass of the C2H2 in the original mixture.

Answer:

C3H8 + C2H2 + O2 ------> CO2 + H2O unbalanced

break the equation down into 2 equations - combustion of c3h8:
C3H8 + 5O2 ---> 4H2O + 3CO2

and combustion of C2H2:
2C2h2 + 5O2 ---> 2H2O + 4CO2

'integrate' balanced equations into full equation:

C3H8 + 2C2H2 + 10O2 ---> 6H2O + 7CO2

you need 1.5 as many moles of co2 than water, and the only way you can achieve this is by increasing the moles of C2H2 (as this individual reaction produces more CO2 moles than water moles)

if we multiply the combustion of C2H2 by 2:

4C2H2 + 10O2 ---> 4H2O + 8CO2

then add this onto our full equation we get:

C3H8 + 6C2H2 + 20O2 ---> 10H2O + 15CO2

6 moles C2H2 x RMM (26) = 156g
1 mole C3H8 x RMM (44) =44g

so C3H8 and C2H2 react in the ratio
44:156
multiply by 0.01:
0.44g : 1.56g
1.56 ~ 1.6g