balanced equation is
2CaO + 5C ------------>2 CaC2 + CO2
75.5g/56g/mol 50g/12g/mol 0 0 intital moles
=1.348 =4.167
To know the limiting reagent , let us calculate the ratio of given to required moles
1.348/2= 0.674 4.167/5=0.833
Thus CaO is limiting reagent and C is excess reactant
Thus after reaction
2CaO + 5C ------------> 2 CaC2 + CO2
75.5g/56g/mol 50g/12g/mol 0 0 intital moles
=1.348 =4.167
0 4.167- ( 1.348x5/2)=2.482 1.348 0.674 after reaction
thus the excess reactant (carbn) remaining = 2.482 moles
mass remaining of Carbon = molesx molar mass= 2.482 mol x 12g/mol =29.784 g
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