M(P2O5) = 8 грамм m(H2O) = 10 грамм H2O+ P2O5=H3PO4 m(H3PO4) - ?

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вопрос

To solve this problem, we need to find the number of moles of each substance first.

Calculate the number of moles of P2O5
Molar mass of P2O5 = 2(30.97) + 5(16.00) = 141.94 g/mo
Number of moles of P2O5 = 8 g / 141.94 g/mol ≈ 0.056 moles

Calculate the number of moles of H2O
Molar mass of H2O = 2(1.01) + 16.00 = 18.02 g/mo
Number of moles of H2O = 10 g / 18.02 g/mol ≈ 0.555 moles

The balanced chemical equation for the reaction is
P2O5 + 3H2O → 2H3PO4

Use the mole ratio in the balanced chemical equation to determine the limiting reactant and the moles of H3PO4 formed
From the reaction, 1 mole of P2O5 reacts with 3 moles of H2O to produce 2 moles of H3PO4.

Since we have 0.056 moles of P2O5 and 0.555 moles of H2O, the limiting reactant is P2O5 because it will run out first. Therefore, all 0.056 moles of P2O5 will react.

Calculate the moles of H3PO4 formed
Moles of H3PO4 = 0.056 moles * (2 moles H3PO4 / 1 mole P2O5) = 0.112 moles

Calculate the mass of H3PO4 formed
Molar mass of H3PO4 = 3(1.01) + 30.97 + 4(16.00) = 97.99 g/mo
Mass of H3PO4 = 0.112 moles * 97.99 g/mol ≈ 10.95 grams

Therefore, the mass of H3PO4 formed in the reaction is approximately 10.95 grams.