Chapter 11, Problem 99A

Fertilizer The compound calcium cyanamide $\left(\text{CaNCN}\right)$ is used as a nitrogen source for crops. To obtain this compound, calcium carbide is reacted with nitrogen at high temperatures.
   ${\text{CaC}}_{\text{2}}\left(\text{s}\right){\text{+N}}_{\text{2}}\left(\text{g}\right)\to \text{CaNCN}\left(\text{s}\right)\text{+C}\left(\text{s}\right)$

What mass of $\text{CaNCN}$ can be produced if 7.50 mol of ${\text{CaC}}_{\text{2}}$ reacts With 5.00 molOf ${\text{N}}_2$ ?

Interpretation Introduction

Interpretation:

Mass of CaNCN needs to be determined.

Concept introduction:

Calcium cyanamide is used as a nitrogen source for the crops.

It is produced by the reaction of Calcium carbide and nitrogen at high temperature.

Answer to Problem 99A

Yield of CaNCN is 400.5 g

Explanation of Solution

Step 1: write balance chemical equation

$\begin{array}{l}CaC2(s)+N2(g)\to CaNCN(s)+C(s)\\ 7.50moles5moles?\end{array}$

Step 2: Convert moles of reactant to moles of product using mole ratio

1 mole of CaC₂ produce = 1 moles of CaNCN

7.5 moles of CaC₂ produce = 7.5 moles ofCaNCN

1 mole of N₂ produce = 1 moles of CaNCN

5 moles of N₂ produce = 5 moles of CaNCN

N₂ produce less number of moles of CaNCN. So, it acts as limiting reactant.

Step 3: Convert moles of product to mass (theoretical yield)

$\begin{array}{c}Molar\text{}massofCaNCN=80.10g/mole\\ Mass=Molarmass\times Moles\\ MassofCaNCN(Theoreticalyield)=80.10\times 5=400.5g\end{array}$