Question

If 5.00 g of pure Na2CO3 is treated with HCL solution, how many grams of NaCl will be produced if the water is subsequently evaporated from the salt?

Answer 1

5.499 g of NaCl is formed if 5.00 g of pure Na2CO3 is treated with HCl solution

The equation of the reaction is;

2HCl  +  Na2CO3 ----->2NaCl + H2O + CO2

Molar mass of Na2CO3 = 2(23) + 12 + 3(16) = 46 + 12 +48 = 106 g/mol

The number of moles of Na2CO3 in 5.00 g of pure Na2CO3 = 5g/106 g/mol

= 0.047 moles

From the reaction equation;

1 mole of Na2CO3  yields 2 moles of NaCl

0.047 moles of Na2CO3  yields 0.047 × 2/1 = 0.094 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl = 0.094 moles × 58.5 g/mol

Mass of NaCl =  5.499 g of NaCl

Learn more: brainly.com/question/9743981