Question

448 L of ammonia at STP on decomposition gives X g of nitrogen. How many molecules of oxygen are required to produce NO₂ from X g of nitrogen.

Answer 1

Step-1: Calculate the value of X g of nitrogen

$$\begin{gathered} 2{\mathrm{NH}}_3\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{Ammonia}\right)\left(\mathrm{Nitrogen}\right)\left(\mathrm{Hydrogen}\right) \end{gathered}$$

According to Avogadro’s law, at STP, 22.4 L of any gas weighs equal to the gram molecular weight of the substance.

The gram atomic weight of Nitrogen (N)= 14 g mol^-1

The gram atomic weight of Hydrogen (H)= 1 g mol^-1

Gram molecular weight of Ammonia (NH₃)= 14+3(1)= 17 g mol^-1

Therefore, 22.4 L of Ammonia (NH₃)= 17 g of Ammonia

From the balanced chemical equation,

2× 22.4 L of ammonia gas decomposes to form= 28 g of N₂

1 L of ammonia gas decomposes to form= $\frac{28}{2\times 22.4}$ g of N₂

448 L of ammonia gas decomposes to form= $\frac{28}{2\times 22.4}\times 448=280$ g of N₂

Step-2: Calculate the weight of oxygen required to react with 280 g of nitrogen to produce NO₂

$$\begin{gathered} 2{\mathrm{O}}_2\left(\mathrm{g}\right)+{\mathrm{N}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NO}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{Oxygen}\right)\left(\mathrm{Nitrogen}\right)(\mathrm{Nitrogen}\mathrm{dioxide}) \end{gathered}$$

From the balanced chemical equation,

28 g of nitrogen reacts with= 64 g of oxygen

1 g of nitrogen reacts with= $\frac{64}{28}$ g of oxygen

280 g of nitrogen reacts with= $\frac{64}{28}\times 280=640$ g of oxygen

Step-3: Calculate the number of molecules of oxygen

According to the mole concept, the gram molecular weight of a substance= 6.023×10²³ molecules per mole

Therefore, 32 g of oxygen= 6.023×10²³ molecules

1 g of oxygen= $\frac{6.023x{10}^{23}}{32}$ molecules

640 g of oxygen= $\frac{6.023x{10}^{23}}{32}\times 640=12\times {10}^{24}$molecules

Therefore, the molecules of oxygen required to produce NO₂ from X g of nitrogen are 12 ×10²⁴ molecules.