This question has multiple parts. Work all the parts to get the most points. For

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This question has multiple parts. Work all the parts to get the most points. For the following unbalanced chemical equation, suppose that exactly 200 g of each reactant is taken. what mass of CO2 is expected (assuming that the limiting reactant is completely consumed). Determine which reactant is limiting, and calculate CS2(1)02(CO2(9)S02 (9) Limiting reactant Mass of CO2 b For the following unbalanced chemical equation, suppose that exact?ly 1.30 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass ofH2O is expected (assuming that the limiting reactant is completely consumed). NH, (g) + CO2 (g) CN2H40(s) + H2O(1) Limiting reactant: Mass of H20 GFor 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 (assuming that the limiting reactant is completely consumed). H2(g) + MnO2(s) -Mno(s) H2O(l) Limiting reactant: Previous N

Explanation / Answer

a) balanced equation : CS2(l) + 3O2(g) ----> CO2(g) + 2SO2(g)

No of mol of CS2 = 2/76 = 0.0263 mol

No of mol of O2 = 2/32 = 0.0625 mol

1 mol CS2 = 3 mol O2

limiting reactant = O2

No of mol of CO2 formed = 0.0625*1/3 = 0.0208 mol

mass of co2 = 0.0208*44 = 0.915 g

b)
balanced equation : 2NH3(g) + CO2(g) ----> CN2H4O(s) + H2O(l)

no of mol of NH3 = W/mWT = 1.3/17 = 0.0765 mol

NO of mol of CO2 = 1.3/44 = 0.0295 mol

Limiting reactant = Co2

mass of H2O = 0.0295*1/1*18 = 0.531 g

C) balanced equation : H2(g) + MnO2(s) ----> MnO(s) + H2O(l)

no of mol of H2 = 1.4/2 = 0.7 mol

no of mol of MnO2 = 1.4/87 = 0.0161 mol

limiting reactant = MnO2

mass of H2O = 0.0161*18 = 0.29 g

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