Question

multiple choice 10 points
calcium carbide (cac₂) is prepared as per the following reaction:
cao + 3c → cac₂ + co
determine how much of it may be produced, in lb, if 0.557 lb of cao and 0.557 lb of c react.
0.212 lb
0.636 lb
0.557 lb
0.992 lb

Explanation:

Step1: Calculate moles of reactants

Molar mass of $CaO$ is approximately $56.08\ g/mol$, molar mass of $C$ is $12.01\ g/mol$. First convert pounds to grams ($1\ lb = 453.592\ g$).
Moles of $CaO=\frac{0.557\times453.592}{56.08}\approx4.52\ mol$
Moles of $C=\frac{0.557\times453.592}{12.01}\approx21.0\ mol$

Step2: Determine limiting reactant

From the balanced equation $CaO + 3C
ightarrow CaC_{2}+CO$, the mole - ratio of $CaO$ to $C$ is $1:3$.
For $4.52\ mol$ of $CaO$, we need $4.52\times3 = 13.56\ mol$ of $C$. Since we have $21.0\ mol$ of $C$, $CaO$ is the limiting reactant.

Step3: Calculate moles of $CaC_{2}$ produced

The mole - ratio of $CaO$ to $CaC_{2}$ is $1:1$. So, moles of $CaC_{2}$ produced is equal to moles of $CaO$ which is $4.52\ mol$.

Step4: Calculate mass of $CaC_{2}$ produced

Molar mass of $CaC_{2}$ is $40.08 + 2\times12.01=64.1\ g/mol$.
Mass of $CaC_{2}$ in grams is $4.52\times64.1 = 290.732\ g$.
Convert to pounds: $\frac{290.732}{453.592}\approx0.636\ lb$

Answer:

$0.636\ lb$