Question

In the reaction 2 Cr+2 H3PO4→3 H2+2 CrPO42 Cr+2 H3PO4→3 H2+2 CrPO4, if 118.5 mL of a 0.8775 M solution of H3PO4 is used, what is the number of moles and mass of H2 produced?

a) Moles of H2 = 0.8775 mol; Mass of H2 = 0.8775 g
b) Moles of H2 = 1.755 mol; Mass of H2 = 1.755 g
c) Moles of H2 = 0.439 mol; Mass of H2 = 0.439 g
d) Moles of H2 = 0.2195 mol; Mass of H2 = 0.2195 g
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Answer 1

In the reaction 2Cr + 2H₃PO₄ → 3 H₂ + 2CrPO₄, if 118.5 mL of a 0.8775 M solution of H₃PO₄, the number of moles and mass of H₂ produced is Moles of H₂ = 0.1559 mol; Mass of H₂ = 0.3118 g ( none of the given options can be selected as the correct answer.)

Step-by-step explanation:

To determine the number of moles and mass of H₂ produced in the given reaction, we need to use the provided information.

First, we convert the volume of the H₃PO₄ solution to moles using its molarity:

0.8775 M * 0.1185 L = 0.1039 mol H₃PO₄

From the balanced equation, we can see that the mole ratio between H₂ and H₃PO₄ is 3:2.

Therefore, the number of moles of H₂ produced is:

0.1039 mol H₃PO₄ * (3 mol H₂ / 2 mol H₃PO₄) = 0.1559 mol H₂

The molar mass of H₂ is 2 g/mol, so the mass of H₂ produced is:

0.1559 mol H₂ * 2 g/mol = 0.3118 g H₂

Therefore, the correct answer is:

Moles of H₂ = 0.1559 mol; Mass of H₂ = 0.3118 g. Therefore, none of the given options can be selected as the correct answer.