Question

How many grams of water form when 4.80L of propane gas (C3H8) completely react?

How many grams of CO2 are produced from 18.2g of oxygen gas and excess propane?

How many grams of H2O can be produced from the reaction of 8.30

Answer 1

Answer 1   C3H8 + 5O2 ==> 3CO2 + 4H2O

When using gases at the same T and P, one can take a shortcut and use liters as if they were mols. For the problem we simply convert L of one thing into L of the other.
0.7L C3H8 x (5 mols O2/1 mol C3H8) = ? L O2 required.

Answer 2 0.990g

Answer 3 First balance the equation between H2O and C3H8

C3H8 + 5O2 -------> 3CO2 + 4H2O

We will use the balanced equation later, but for now we need to find the moles of C3H8 by dividing the number of molecules of C3H8 by avogadros number:
(8.50 X 10^22 molecules of C3H8)/(6.023x10^(23) molecules/mole) = 0.141 moles of C3H8

Now that we have the number of moles of C3H8 we need to use the mole ratio between H2O and C3H8 by looking at the coefficients above
(0.141 moles of C3H8)x(4 moles of H2O)/(1 mole of C3H8) = 0.565 moles of H2O

We now have the moles of H2O, so multiply the number of moles of H2O by the molar mass of H2O
(0.565 moles of H2O)(18.02 grams/mole of H2O) = 10.2 grams of H2O.
Let me know if you need me to clarify on anything.