Question

17 g of ammonia is completely reacted with 45 g of oxygen to produce NO and H2O (water). which is limiting reagent? calculate the number of moles of unreacted reactant left over. what volume of NO are produce at NTP ? calculate the mass of water produce.

Answer 1

The mass of water produced is  18 g.

To determine the limiting reagent, we need to compare the number of moles of each reactant.

First, we need to convert the given masses of ammonia and oxygen to moles:

Moles of ammonia = $17 g / 17 g/mol = 1 mol$

Moles of oxygen = $45 g / 32 g/mol = 1.41 mol$

From the balanced equation, it can be seen that 1 mole of ammonia reacts with 1 mole of oxygen to produce 1 mole of NO and 1 mole of H2O (water). Since there are more moles of oxygen available than ammonia, we can conclude that the limiting reagent is ammonia.

Now, we can use the stoichiometry of the reaction to find the number of moles of unreacted reactant left over:

1 mol of ammonia reacts with 1.41 mol of oxygen, so the number of moles of oxygen left over is 1.41 mol - 1 mol = 0.41 mol

To calculate the volume of NO produced at NTP (Standard Temperature and Pressure) :

We know that 1 mole of NO is produced for each mole of ammonia reacted. So, the number of moles of NO produced is 1 mol.

NO is a gas, and at NTP, its molar volume is 24.4 L/mol.

Therefore, the volume of NO produced is $24.4 L/mol * 1 mol = 24.4 L.$

To calculate the mass of water produced:

We know that 1 mole of H2O is produced for each mole of ammonia reacted. So, the number of moles of H2O produced is 1 mol.

The molar mass of water is 18 g/mol.

Therefore, the mass of water produced is $18 g/mol * 1 mol = 18 g.$

To know more about the concept please go through the links

brainly.in/question/29730778

brainly.in/question/8823627

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