CaO + 3C = CaC2 + CO What is the amount of CaC2 (in g) which can be produced from the reaction of excess calcium oxide and 4.1 g of carbon?
Q174322
CaO + 3C = CaC2 + CO
What is the amount of CaC2 (in g) which can be produced from the reaction of excess calcium oxide and 4.1 g of carbon?
Solution:
The given reaction is already balanced for all the elements.
Calcium oxide and carbon are the reactants.
Calcium oxide is in excess. So the amount of products formed would depend on the mass of carbon.
Carbon would be the Limiting Reactant and we can find the mass of products formed using the mass of carbon.
Step 1: Convert 4.1 grams of carbon to moles.
Atomic mass of carbon = 12.011 g/mol
"= 0.3414\\space mol \\space of\\space C"
Step 2: Find moles of calcium carbide ( CaC2 ) using moles of carbon.
The mole to mole ratio of Carbon and calcium carbide in the given reaction is
= 3 mol of C : 1 mol of CaC2.
Using this ratio find the moles of calcium carbide that can be formed from 0.3414 mol of C.
"= 0.1138\\space mol \\space of\\space CaC_2"
Step 3: Convert 0.1138 mol of CaC2 to grams by using the molar mass of CaC2.
atomic mass of Ca = 40.078 g/mol , atomic mass of C = 12.011 g/mol
molar mass of CaC2 = 1 * atomic mass of Ca + 2 * atomic mass of C
= 1 * 40.078 g/mol + 2 * 12.011 g/mol
= 40.078 g/mol + 24.022 g/mol
= 64.1 g/mol
"= 7.295\\space grams\\space of\\space CaC_2"
In the question, we are given the mass of carbon in 2 significant figure, so our final answer must also be
in 2 significant figures.
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