Question

How many miligrams of CO2 and H2O are produced from the complete combustion of 5.650 mg of C10H7OH?

Answer 1

The reaction is as follow:
C₁₀H₇OH + O₂ ---------> CO₂ + H₂O

And the balanced reaction is (the order I use was first the C, the H and the last, the oxygen)
2C₁₀H₇OH + 23O₂ ---------> 20CO₂ + 8H₂O

Or you can use this alternate balance reaction:
C₁₀H₇OH + 23/2O₂ ---------> 10CO₂ + 4H₂O

You will get the same result no matter which you use. In this case, I'm gonna use the second reaction. LEt's get the moles of C₁₀H₇OH and then the moles of CO₂ and H₂O by stechiometry:

moles C₁₀H₇OH = 5.650x10^-3 g / (10*12 + 8*1 + 16) = 3.92x10^-5 moles

By stechiometry we produce 10 moles of CO2 and 4 of water, so:
moles CO₂ = 3.92x10^-5 * 10 = 3.94x10^-4 moles
mass CO₂ = 3.94x10^-4 moles * (12+32) = 0.01725 g ---> 17.25 mg

moles H₂O = 3.92x10^-5 * 4 = 1.576x10^-4 moles
mass H₂O = 1.576x10^-4 * (2+16) = 0.00284 g ---> 2.84 mg

Hope this helps