Calcium cyanamide reacts with H₂O(g) at 150^$\omicron$C and 1.00 atm pressure:
CaNCN(s) + 3H₂O(g) $\rightarrow$ CaCO₃(s) + 2NH₃(g)
If 80.0 L of NH₃(g) Are produced, What volume of steam is required?

Question

Calcium cyanamide reacts with H₂O(g) at 150^$\omicron$C and 1.00 atm pressure:
CaNCN(s) + 3H₂O(g)

\rightarrow

CaCO₃(s) + 2NH₃(g)
If 80.0 L of NH₃(g) Are produced, What volume of steam is required?

Quizplus · Accepted Answer

The balanced chemical equation shows that for every 2 moles of $NH_3$ produced, 3 moles of $H_2O$ are required. Since 80.0 L of $NH_3$ are produced, the volume of steam required is $80.0 \,L \times \frac{3}{2} = 120.0 \,L$, assuming the reaction occurs under the same conditions of temperature and pressure (ideal gas behavior).

Calcium Cyanamide Reacts with H2O(g) at 150 ο\omicron ο C and 1

Calcium Cyanamide Reacts with H₂O(g) at 150^$\omicron$C and 1