Question

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How many moles of P2O3 are required to fully react with 108 H2O? (H2O; 18 g/mol)
P2O3 + 3H2O --> 2H3PO3
108 gH2O ---> mol P2O3

Answer 1

108 g H2O x (1 mol H2O / 18 g H2O) = 6 moles H2O

According to the balanced chemical equation, 3 moles of H2O react with 1 mole of P2O3 to produce 2 moles of H3PO3. Therefore, you need to divide the number of moles of H2O by 3 and multiply by 2 to get the number of moles of H3PO3 required:

6 moles H2O / 3 x 2 moles H3PO3 / 1 mole P2O3 = 4 moles P2O3

So, you would need 4 moles of P2O3 to fully react with 108 g of H2O.

Answer 2

To determine the moles of P2O3 required to react with 108 g H2O, we calculate the moles of H2O and then use the stoichiometry of the balanced chemical equation. We find that 2 moles of P2O3 are needed.

Step-by-step explanation:

To find out how many moles are present in 108 grams of water (H2O), we use the molecular weight of H2O which is 18.02 g/mol as given.

Using the formula:
Number of moles = mass (g) / molecular weight (g/mol),
we get the moles of H2O: 108 g / 18.02 g/mol.

We find that there are 6 moles of H2O. Now, looking at the balanced chemical equation P2O3 + 3H2O → 2H3PO3, we see that 3 moles of water react with 1 mole of P2O3. Therefore, to react with 6 moles of H2O, we would need 2 moles of P2O3.