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)

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