Ask Question
13 June, 12:40

How many grams of sodium carbonate would be required to produce 85.0 g of barium carbonate? (the molar mass of Na2CO3 is 106 g per mole, and the molar mass of BaCO3 is 197 grams per mole)

+1
Answers (1)
  1. 13 June, 12:51
    0
    Mass of sodium carbonate required = 45.58 g

    Explanation:

    Given dа ta:

    Mass of sodium carbonate required = ?

    Mass of barium carbonate produced = 85.0 g

    Solution:

    Chemical equation:

    Na₂CO₃ + BaCl₂ → BaCO₃ + 2NaCl

    Number of moles of BaCO₃:

    Number of moles = mass / molar mass

    Number of moles = 85.0 g / 197 g/mol

    Number of moles = 0.43 mol

    Now we will compare the moles of sodium carbonate with barium carbonate.

    BaCO₃ : Na₂CO₃

    1 : 1

    0.43 : 0.43

    Mass of Na₂CO₃:

    Mass = number of moles * molar mass

    Mass = 0.43 mol * 106 g/mol

    Mass = 45.58 g
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “How many grams of sodium carbonate would be required to produce 85.0 g of barium carbonate? (the molar mass of Na2CO3 is 106 g per mole, ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers