Answer:
a) the limiting reactant is 02
b) There will remain 0.667 moles of CS2
c) There will be formed 0.333 moles oof CO2 and 0.667 moles of SO2
Explanation:
Step 1: Data given
Number of moles of CS2 = 1.00 mol
Number of moles of O2 = 1.00 mol
Molar mass of O2 = 32 g/mol
Molar mass of CS2 = 76.14 g/mol
Step 2: The balanced equation
CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)
Step 3: Calculate the limiting reactant
For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2
O2 is the limiting reactant. It will completely be consumed.(1.00 mol).
CS2 is in excess. There will react 1.00/ 3 = 0.333 moles
There will remain 1.00 - 0.333 = 0.667 moles of CS2
Step 4: Calculate moles of CO2 and SO2
For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2
For 1.00 mol of O2 we have 1.00/3 = 0.333 moles CO2
For 1.00 mol of O2 we have 1.00 /(3/2) = 0.667 moles of SO2
