During an experiment, 104 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. if the percent yield of the reaction was 80.15%, what was the actual amount of calcium chloride formed? caco3 + hcl → cacl2 + co2 + h2o 90.1 grams 92.4 grams 109.2 grams 115.3 grams

92.4 grams.

Explanation:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,

1.0 mole of CaCO₃ reacts with 2.0 moles of HCl to produce 1.0 mole of CaCl₂, 1.0 mole of CO₂, and 1.0 mole of H₂O.

no. of moles of CaCO₃ = mass/molar mass = (104 g)/(100.08 g/mol) = 1.039 mol.

Using cross multiplication:

1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.

∴ 1.039 mole of CaCO₃ produce → 1.039 mole of CaCl₂.

∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.039 mol)(110.98 g/mol) = 114.3 g.

∵ percent yield of the reaction = [(actual yield)/(theoretical yield)] x 100.

Percent yield of the reaction = 80.15%, theoretical yield = 115.3 g.

∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100 = [(80.15%)/(115.3 g)] / 100 = 92.42 g ≅ 92.4 g.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :