Determine how many grams of water can be produced from excess propane C3H8 and 2.55g oxygen? MM 02 = 32 and MM H2O = 18.2.

I tried 2.5 gm O2 X 5 mole O2/32.00 gm O2 x 4 mole H2O/5 mole O2 = 1.40. This is close but not correct. The answer is 1.15. Can you help with this?

balance the reaction:

C3H8+O2>> 3CO2 + 4H2O

so, you get 4/5 mole water for each mole of O2.
figure moles of O2: 2.55/32
then moles of H2O: 4/5 * 2.55/32

then grams of water: moleswater*18
Yes, I have rounded off molemasses, you do it more accurately.

I get 1.15 grams.

To determine the grams of water produced from excess propane and 2.55g of oxygen, we need to first balance the chemical equation for the combustion of propane:

C3H8 + 5O2 -> 3CO2 + 4H2O

The balanced equation tells us that for every 1 mole of propane (C3H8), 4 moles of water (H2O) are produced.

Now let's calculate the moles of oxygen (O2) and propane (C3H8) given the mass provided:

Mass of O2 (given) = 2.55g
Molar mass of O2 (MM O2) = 32.00 g/mol

Moles of O2 = Mass of O2 / MM O2 = 2.55g / 32.00 g/mol = 0.0797 mol O2

Since oxygen is in excess, we assume that all the propane reacts completely.

Now, let's calculate the moles of water produced from the moles of propane:

Moles of propane (C3H8) = Moles of O2 (as C3H8 : O2 ratio is 1 : 5)
= 0.0797 mol O2 * (1 mol C3H8 / 5 mol O2) = 0.01594 mol C3H8

Moles of water (H2O) = Moles of propane * (4 mol H2O / 1 mol C3H8)
= 0.01594 mol C3H8 * (4 mol H2O / 1 mol C3H8) = 0.06376 mol H2O

Finally, let's calculate the mass of water produced:

Molar mass of water (MM H2O) = 18.2 g/mol

Mass of water (H2O) = Moles of water * MM H2O
= 0.06376 mol H2O * 18.2 g/mol = 1.16083 g H2O

The mass of water produced from the given information is approximately 1.16g, which is close to the correct answer of 1.15g.

Sure! Let's walk through the steps to determine the correct answer.

First, we need to determine the balanced chemical equation for the reaction between propane (C3H8) and oxygen (O2) to form water (H2O). The balanced equation is:

C3H8 + 5O2 -> 3CO2 + 4H2O

From the balanced equation, we can see that for every 1 mole of C3H8, we need 5 moles of O2 to produce 4 moles of water (H2O).

Next, we need to calculate the number of moles of oxygen we have. Given that the mass of oxygen (O2) is 2.55 grams and the molar mass of oxygen (O2) is 32 g/mol, we can use the formula:

Moles = Mass / Molar mass

Moles of O2 = 2.55 g / 32 g/mol = 0.0797 mol O2

Now, we can use the mole ratio from the balanced equation to find the number of moles of water (H2O) produced. From the balanced equation, we know that 4 moles of H2O are produced for every 5 moles of O2:

Moles H2O = 0.0797 mol O2 * 4 mol H2O / 5 mol O2 = 0.0638 mol H2O

Finally, to calculate the mass of water (H2O) produced, we need to multiply the number of moles of water (H2O) by the molar mass of water (H2O), which is 18.2 g/mol:

Mass H2O = 0.0638 mol H2O * 18.2 g/mol = 1.16 g H2O

Therefore, the correct answer is approximately 1.16 grams of water (H2O) produced.

It seems like there might have been a slight rounding error in your calculations, which could explain the discrepancy between your answer and the correct answer of 1.15 grams.