Question

This question has multiple parts. Work all the parts to get the most points. a For the following unbalanced chemical equation, suppose that exactly 1.85 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of CO2 is expected (assuming that the limiting reactant is completely consumed). CS2() + O2(g) + CO2(g) + SO2(9) Limiting reactant Mass of CO2 = b For the following unbalanced chemical equation, suppose that exactly 1.40 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of H2O is expected (assuring that the limiting reactant is completely consumed). NH: (9) + CO2(g) → CN, H40(s) + H2O(1) Limiting reactant Mass of H2O =

Answer 1

a) Balanced eequation: CS2 302 > CO2 + 2SO2 Number of moles of CS2 mass/molar mass = 1.85g/76.1g/mol =0.0243 mol Number of moles of O2 = mass/molar mass = 1.85g/32.0g/mol 0.0578 mol Thus the O2 is in excess and hence Limiting reagent: CS2 Number of moles of CO2 expected Mass of 0.0243mol of CO2 0.0243 moles mole*molar mass =0.0243 44.0 g/mol = 1.07 g Mass of CO2 = 1.07 g

b) Balanced eequation: 2NH3 CO2 -> CN2H40 + H2O Number of moles of NH3 = 1.40g/17.0g/mol =0.0824 mol Number of moles of CO2 = 1.40g/44.0g/mol 0.0318 mol Thus the NH3 is in excess and hence Limiting reagent: CO2 Number of moles of H20 expected 0.0318 moles Mass of 0.0318mol of H20 =0.0318 18.0 g/mol 0.572 g Mass of H20 0.572 g c) Balanced eequation: H2MnO2> MnO Number of moles of H2 H2o = 1.10g/2.02g/mol 0.545 mol Number of moles of MnO2 = 1.10g/86.9g/mol =0.0127 mol Thus the H2 is in excess and hence Limiting reagent: MnO2 Number of moles of H20 expected 0.0127 moles Mass of 0.0127mol of H20 =0.0127*18.0 g/mol 0.229 g Mass of H20 = 0.229 g

d) Balanced eequation: 12CI2>2ici = 1.25g/254g/mol 0.00492 mol = 1.25g/70.9g/mol =0.0176 mol Number of moles of 12 Number of moles of C12 Thus the C12 is in excess and hence Limiting reagent: 12 Number of moles of ICl expected 0.00492*2 moles =0.00984 =0.00984*162 g/mol =1.59 g Mass of 0.00984mol of ICI Mass of ICI 1.59 g