For the following reaction, 5.62 grams of oxygen gas are mixed with excess ammonia. The reaction yields 3.58 grams of nitrogen monoxide. ammonia (g) oxygen (g) nitrogen monoxide (g) water (g) What is the ideal yield of nitrogen monoxide

The theoretical yield of nitrogen monoxide is 4.23 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of oxygen gas = 5.62 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of oxygen gas and ammonia is:

By Stoichiometry of the reaction:

5 moles of oxygen gas produces 4 moles of nitrogen monoxide

So, 0.176 moles of oxygen gas will produce = moles of nitrogen monoxide

Now, calculating the mass of nitrogen monoxide from equation 1, we get:

Molar mass of nitrogen monoxide = 30 g/mol

Moles of nitrogen monoxide = 0.141 moles

Putting values in equation 1, we get:

Hence, the theoretical yield of nitrogen monoxide is 4.23 grams