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In the reaction 2Fe + 3Cl2 β†’ 2FeCl3, how many grams of FeCl3 can be produced fro

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Question: In the reaction 2Fe + 3Cl2 β†’ 2FeCl3, how many grams of FeCl3 can be produced from 10 g of Fe?

Options:

  1. 20 g
  2. 30 g
  3. 40 g
  4. 50 g

Correct Answer: 40 g

Solution:

10 g of Fe = 0.18 moles. 2 moles of Fe produce 2 moles of FeCl3. 0.18 moles of FeCl3 = 0.18 * 162.5 g = 29.25 g.

In the reaction 2Fe + 3Cl2 β†’ 2FeCl3, how many grams of FeCl3 can be produced fro

Practice Questions

Q1
In the reaction 2Fe + 3Cl2 β†’ 2FeCl3, how many grams of FeCl3 can be produced from 10 g of Fe?
  1. 20 g
  2. 30 g
  3. 40 g
  4. 50 g

Questions & Step-by-Step Solutions

In the reaction 2Fe + 3Cl2 β†’ 2FeCl3, how many grams of FeCl3 can be produced from 10 g of Fe?
  • Step 1: Calculate the number of moles of Fe in 10 grams. Use the formula: moles = mass (g) / molar mass (g/mol). The molar mass of Fe is approximately 55.85 g/mol.
  • Step 2: Substitute the values into the formula: moles of Fe = 10 g / 55.85 g/mol.
  • Step 3: Perform the calculation to find the number of moles of Fe. This gives you approximately 0.18 moles of Fe.
  • Step 4: Look at the balanced chemical equation: 2Fe + 3Cl2 β†’ 2FeCl3. This shows that 2 moles of Fe produce 2 moles of FeCl3.
  • Step 5: Since 2 moles of Fe produce 2 moles of FeCl3, 0.18 moles of Fe will produce the same amount of FeCl3, which is also 0.18 moles of FeCl3.
  • Step 6: Calculate the mass of FeCl3 produced. The molar mass of FeCl3 is approximately 162.5 g/mol.
  • Step 7: Use the formula: mass = moles Γ— molar mass. Substitute the values: mass of FeCl3 = 0.18 moles Γ— 162.5 g/mol.
  • Step 8: Perform the calculation to find the mass of FeCl3 produced, which is approximately 29.25 grams.
  • Stoichiometry – The calculation of reactants and products in chemical reactions based on balanced equations.
  • Mole Concept – Understanding the relationship between grams, moles, and molar mass in chemical calculations.
  • Molar Mass Calculation – Calculating the mass of a compound using its molecular formula and the atomic masses of its constituent elements.
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