Question

Assuming that pressure and temperature are constant, how many liters of oxygen gas will be required to produce 0.160 L of nitrogen gas.

4 NH3 (g) + 3 O2 (g) → 6 H2O (g) + 2 N2 (g)

Answer 1

Answer:

0.24 L of oxygen gas will be required to produce 0.160 L of nitrogen gas.

Explanation:

From the reaction, 3 moles of oxygen react with ammonia to produce 2 moles of nitrogen gas

So, 3 moles of 02 = 2 moles of N2

Since I mole of a gas occupies 22.4dm^3 or 22,4 L of the gas

3 * 22,4L of O2 produces  2 * 22.4 L of N2

67.2 L of O2 produces 44.8 L of N2

( 67.2 * 0.160 / 44.8 ) L of O2 will produce 0.160L of N2

10.752/ 44,8 L = 0.24 L of O2 will produce 0.160 L of N2

0.24 L of oxygen gas will be required to produce 0.160 L of nitrogen gas.

Answer 2

Molar mass is defined as the ratio of the given mass of a compound to the number of moles present in that compound. The molar mass has the SI unit in grams.

The 0.24 L of oxygen gas will be required to produce 0.160 L of nitrogen gas.

From the chemical equation given, it can be interprested as:

4 NH₃ (g) + 3 O₂ (g) → 6 H₂O (g) + 2 N₂ (g)

3 moles of oxygen are required to produce 2 moles of ammonia.

That is:

3 moles of oxygen = 2 moles of nitrogen

Now, the 1 moles of gas occupies the 22.4 L of the gas, such that:

3 x 22.4 L of oxygen = 2 x 22.4 L of Nitrogen

67.2 L of oxygen = 44.8 L of nitrogen.

Now, given the 0.160 L of nitrogen gas-producing oxygen will be:

$O_2&=\dfrac{67.2 \times 0.160}{44.8 }\\\\\\O_2&=\dfrac{10.752}{44.8 }$

Thus, 0.24 L of oxygen will be required to produce 0.160 L of nitrogen gas.

To know more about moles and molar mass, refer to the following link:

https://brainly.com/question/16386473