Answer :
Answer:
1,1314.56 grams of calcium carbide can be prepared.
Explanation:
The balanced chemical reaction ;
[tex]2C+CaO\rightarrow CaC_2+CO[/tex]
Mass of carbon present = 1.15 kg = 1150 g( 1 kg = 1000 g)
Moles of carbon = [tex]\frac{1150 g}{12 g/mol}=95.83 mol[/tex]
Mass of CaO present = 1.15 kg = 1150 g
Moles of CaO= [tex]\frac{1150 g}{56g/mol}=20.54 mol[/tex]
According to reaction, 2 moles of carbon reacts with 1 mole of CaO. Then 95.83 moles of carbon will react with ;
[tex]\frac{1}{2}\times 95.83 mol=47.915 mol[/tex] of CaO.
According to reaction, 1 moles of CaO reacts with 2 mole of carbon . Then 20.54 moles of carbon will react with ;
[tex]\frac{2}{1}\times 20.54 mol=41.08 mol[/tex] of carbon
This means that carbon is in excess and CaO is in limiting amount ,so CaO becomes limiting reagent and amount of calcium carbide will depend upon amount of CaO.
According to reaction , 1 mole of CaO gives 1 mole of calcium carbide.Then 20.54 moles of CaO will giv e:
[tex]\frac{1}{1}\times 20.54 mol=20.54 mol[/tex] of calcium carbide
Mass of 20.54 mole of calcium carbide:
20.54 mol × 64 g/mol =1,1314.56 g
1,1314.56 grams of calcium carbide can be prepared.