Question

Excess NaHCO3(s) is added to 535 mL of Cu(NO3)2(aq) 0.240 M for the reaction Cu(NO3)2(aq)+2NaHCO3(s)→CuCO3(s)+2NaNO3(aq)+H2O(l)+CO2(g). a)How many grams of NaHCO3(s) the will be consumed? b)How many grams of CuCO3(s) will be produced?

Answer 1

a) Number of moles of Cu(NO3)2 = Volume of solution (in L) * Molarity (M) = 535/1000 * 0.240 = 0.1284 moles

As per balanced reaction, one mole of Cu(NO3)2 reacts with 2 moles of NaHCO3

Moles of NaHCO3 consumed = 2 * 0.1284 = 0.2568 mol

Molar mass of NaHCO3 = Molar mass of Na + Molar mass of H + Molar mass of C + 3 * Molar mass of O = 23 + 1 + 12 + 3 * 16 = 84 g/mol

Mass of NaHCO3 consumed = number of moles * molar mass = 0.2568 mol * 84 g/mol = 21.57 grams

b) Number of moles of CuCO3 produced = Number of moles of Cu(NO3)2 = 0.1284 moles

Molar mass of CuCO3 = Molar mass of Cu + Molar mass of C + 3 * Molar mass of O = 63.5 + 12 + 3 * 16 = 123.5 g/mol

Mass of CuCO3 produced = number of moles * molar mass = 0.1284 mol * 123.5 g/mol = 15.86 grams

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