Question

Suppose 5.00 g of sodium carbonate reacts completely with an excess of hydrochloric acid. show how to calculate the mass (g) of sodium chloride that would be formed. what is the limiting reagent?

Answer 1

Na2CO3 + 2HCl = 2NaCl + H2O + CO2
n (Na2CO3) = m / M = 5 g / 106 (g/mol) = 0.047 mol
n (NaCl) = 2 * n (Na2CO3) = 0.047 mol * 2 = 0.094 mol
m (NaCl) = n * M = 0.094 mol * 58.5 (g/mol) = 5.5 g
Na2CO3 is the limiting reagent.
Answer:
m (NaCl) = 5.5 g
Na2CO3 is the limiting reagent.