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Leni [432]
3 years ago
14

H2 + NO → H2O + N2 If 180.5 grams of N2 are produced, how many grams of H2 were reacted?

Chemistry
1 answer:
Lena [83]3 years ago
6 0

Answer:

12.89 moles

Explanation:

Before we solve the question, we have to balance the equation of the reaction first. The balanced reaction will be:

2 NO + 2 H2→ N2 + 2 H2O

There are 180.5g of N2 produced, the number of produced in moles will be: 180.5g / (28g/mol)= 6.446 moles

The coefficient of H2 is two and the coefficient of N2 is one. Mean that we need two moles of H2 for every one mole of N2 produced. The number of H2 reacted will be: 2/1 * 6.446 moles = 12.89 moles

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8. 1.167 M

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Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

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8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

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Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

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