Question

Part E Determine the limiting reactant, given 250 g of each reactant, in the following: 3NO2(g) + H2O(1) + 2HNO3(aq) + NO(g) ANSWER: HNO3 Ο NO HO O NO₂ Part F Calculate the grams of HNO3 that would be produced. Express your answer with the appropriate units. ANSWER: m(HNO3) = Part 6 Determine the limiting reactant, given 25.0 g of each reactant, in the following: CHAO(1)+302(g) 2002(g) + 3H2O(g) ANSWER: CO2 O C₂H₂O H2O 02

Answer 1

1. Moles of H₂O = mass / molar mass = 25 / 18 = 1.39 moles

Moles of NO₂ = mass / molar mass = 25 / 46 = 0.544 moles

We see that NO₂ is the limiting reagent

Moles of HNO₃ = moles of NO₂ * 2 /3 = 0.544 * 2 / 3 = 0.363 moles

Mass = moles * molar mass = 0.363 * 159 = 57.72 g

2. Moles of ethanol = mass / molar mass = 25 / 46 = 0.544 moles

Moles of O_{2 =} mass / molar mass = 25 / 32 = 0.781 moles

We se that O₂ is the limiting reagent

Moles of CO₂ = moles of ethanol * 2 = 0.544 * 2 = 1.09 moles

Mass of CO₂ = moles * molar mass = 47.96 g