Question

How many grams of H3PO3 would be produced from the complete reaction of 93.2 g P2O3? P₂O3 + 3H₂O → 2H3PO3 2₂03 | 19 93.2 g P₂03 [?] Which answer choice number goes in the green box? 1) 1 mol P₂03 2) 110 g P2O3 3) 2 mol H3PO3 4) 82 g H3PO3​

Answer 1

2) 110 g P2O3

Step-by-step explanation:

If the green box is in the denominator of a fraction and represents the molar mass of P2O3, then the correct answer choice would be 2) 110 g P2O3. Here’s how you can use this value to solve the problem:

The balanced chemical equation for the reaction between P2O3 and H2O to produce H3PO3 is: P2O3 + 3H2O → 2H3PO3.

From this equation, we can see that 1 mole of P2O3 reacts with 3 moles of H2O to produce 2 moles of H3PO3. So the number of moles of H3PO3 produced is twice the number of moles of P2O3 that reacted.

The molar mass of P2O3 is approximately 110 g/mol. So 93.2 g of P2O3 is equivalent to (93.2 g) / (110 g/mol) = 0.848 mol of P2O3.

Since the number of moles of H3PO3 produced is twice the number of moles of P2O3 that reacted, the number of moles of H3PO3 produced is 0.848 mol × 2 = 1.696 mol.

The molar mass of H3PO3 is (3 × 1.01 g/mol) + (1 × 30.97 g/mol) + (3 × 16.00 g/mol) = 81.99 g/mol. So 1.696 mol of H3PO3 is equivalent to (1.696 mol) × (81.99 g/mol) = 139 g of H3PO3.

So the complete reaction of 93.2 g P2O3 would produce approximately 139 g of H3PO3.