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
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