Question

Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Express your answer to three significant figures. The reaction requires grams of N2.

Answer 1

Answer is: 92.638 g grams of nitrogen must be consumed.

Balanced chemical reaction: CaC₂ + N₂ → CaCN₂ + C.

m(CaCN₂) = 265 g; mass of calcium cyanamide.

M(CaCN₂) = 80.1 g/mol; molar mass of calcium cyanamide.

n(CaCN₂) = 265 g ÷ 80.1 g/mol.

n(CaCN₂) = 3.31 mol; amount of calcium cyanamide.

From balanced chemical reaction: n(CaCN₂) : n(N₂) = 1 : 1.

n(N₂) = 3.31 mol; amount of nitrogen.

m(N₂) = n(N₂) · M(N₂).

m(N₂) = 3.31 mol · 28 g/mol.

m(N₂) = 92.638 g; mass of nitrogen.

Answer 2

Answer : The grams of $N_2$ consumed is, 89.6 grams.

Solution : Given,

Mass of $CaCN_2$ = 265 g

Molar mass of $CaCN_2$ = 80 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate the moles of $CaCN_2$.

$\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles$

The given balanced reaction is,

$CaC_2+N_2\rightarrow CaCN_2+C$

from the reaction, we conclude that

As, 1 mole of $CaCN_2$ produces from 1 mole of $N_2$

So, 3.2 moles of $CaCN_2$ produces from 3.2 moles of $N_2$

Now we have to calculate the mass of $N_2$

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

$\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g$

Therefore, the grams of $N_2$ consumed is, 89.6 grams.