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 data:

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

56 moles of   lieter in chemistry

a is the correct answer