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H2 is produced by the reaction of 120.9 mL of a 0.8677 M solution of H3PO4 according to the following equation.

H2 is produced by the reaction of 120.9 mL of a 0.8677 M solution of H3PO4 according-example-1

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a. Moles of H2 ≈ \(0.1959 \, \text{mol}\)

b. Mass of H2 ≈ \(0.1959 \, \text{mol} \times 2.016 \, \text{g/mol}\)

To solve this problem, we'll follow these steps:

**Given:**

- Volume of H3PO4 solution = 120.9 mL

- Concentration of H3PO4 solution = 0.8677 M

- Balanced chemical equation: \(2 \text{ Cr} + 2 \text{ H3PO4} \rightarrow 3 \text{ H2} + 2 \text{ CrPO4}\)

**a. Number of Moles of H2 Produced:**

- The balanced chemical equation indicates that 2 moles of H3PO4 produce 3 moles of H2.

- Therefore, we first find the moles of H3PO4 using the formula \( \text{moles} = \text{concentration} \times \text{volume} \).

\[ \text{Moles of H3PO4} = 0.8677 \, \text{M} \times 0.1209 \, \text{L} \]

- Now, we use the mole ratio from the balanced equation to find the moles of H2 produced.

\[ \text{Moles of H2} = \frac{3}{2} \times \text{Moles of H3PO4} \]

**b. Mass of H2 Produced:**

- The molar mass of H2 is approximately 2.016 g/mol.

- We use the moles of H2 calculated in part (a) to find the mass of H2.

\[ \text{Mass of H2} = \text{Moles of H2} \times \text{Molar mass of H2} \]

**Calculations:**

a. \[ \text{Moles of H3PO4} = 0.8677 \, \text{M} \times 0.1209 \, \text{L} \]

\[ \text{Moles of H2} = \frac{3}{2} \times \text{Moles of H3PO4} \]

b. \[ \text{Mass of H2} = \text{Moles of H2} \times \text{Molar mass of H2} \]

**Results:**

a. Moles of H2 ≈ \(0.1959 \, \text{mol}\)

b. Mass of H2 ≈ \(0.1959 \, \text{mol} \times 2.016 \, \text{g/mol}\)

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