Through a solution containing 60 g of sodium hydroxide, passed the carbon dioxide obtained by the action of an excess of hydrochloric acid on 200 g of sodium carbonate. what salt and how much was formed?
m (NaOH) = 60 g
m (Na2CO3) = 200 g
m (salt) -?
1) 2HCl + Na2CO3 => 2NaCl + CO2 ↑ + H2O;
2) M (Na2CO3) = 106 g / mol;
M (NaOH) = 40 g / mol;
3) n (Na2CO3) = m (Na2CO3) / M (Na2CO3) = 200/106 = 1.9 mol;
4) n (NaOH) = m (NaOH) / M (NaOH) = 60/40 = 1.5 mol;
5) NaOH + CO2 => NaHCO3 (sodium bicarbonate);
6) M (NaHCO3) = 84 g / mol;
7) n (NaHCO3) = n (NaOH) = 1.5 mol;
8) m (NaHCO3) = n (NaHCO3) * M (NaHCO3) = 1.5 * 84 = 126 g.
Answer: The mass of NaHCO3 is 126 g.
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