hydrocarbon mixture 60% c3h8, 40% CxHy 10 g of mixture was burned with O2 excess to give 29.02 g co2 and 8.8 g h2o. what is CxHy?

Something is wrong with your problem. Either you have copied it wrong or the person making up the problem goofed. A 10 g mixture that is 60% C3H8 will produce 9.8 g H2O alone which leaves no water for the CxHy to produce.

I think the combustion results in 18.8 g H2O... still don't know how to get CxHy though.

To find the value of CxHy in the hydrocarbon mixture, we need to use the given information about the combustion reaction.

Let's start by calculating the moles of CO2 produced:
Molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol

Moles of CO2 = Mass of CO2 / Molar mass of CO2 = 29.02 g / 44.01 g/mol ≈ 0.6597 mol

Since one mole of C3H8 produces three moles of CO2, we can calculate the moles of C3H8 in the mixture:
Moles of C3H8 = (0.6597 mol CO2) / (3 mol CO2 / 1 mol C3H8) = 0.2199 mol C3H8

Now, let's calculate the moles of H2O produced:
Molar mass of H2O = 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol

Moles of H2O = Mass of H2O / Molar mass of H2O = 8.8 g / 18.02 g/mol ≈ 0.488 mol

Since one mole of C3H8 produces four moles of H2O, we can calculate the moles of C3H8 in the mixture:
Moles of C3H8 = (0.488 mol H2O) / (4 mol H2O / 1 mol C3H8) = 0.122 mol C3H8

Now, subtracting the moles of C3H8 from the total moles in the mixture will give us the moles of CxHy:
Moles of CxHy = Total moles - Moles of C3H8 = (0.2199 mol C3H8 + 0.122 mol C3H8) = 0.3429 mol CxHy

Finally, we need to determine the empirical formula of CxHy. Divide the moles of CxHy by the total moles in the mixture and multiply by 100 to find the percentage of each element:

Percentage of C in CxHy = (Moles of CxHy / Total moles) * 100
= (0.3429 mol / (0.2199 mol + 0.122 mol)) * 100
≈ 82.0%

Percentage of H in CxHy = (Moles of CxHy / Total moles) * 100
= (0.3429 mol / (0.2199 mol + 0.122 mol)) * 100
≈ 18.0%

Therefore, the hydrocarbon CxHy is approximately 82% carbon (C) and 18% hydrogen (H).