For the following reaction, 6.74 grams of oxygen gas are mixed with excess ammonia. The reaction yields 4.04 grams of nitrogen monoxide. ammonia (g) + oxygen (g) nitrogen monoxide (g) + water (g) What is the theoretical yield of nitrogen monoxide ? grams What is the percent yield of nitrogen monoxide ? %

The theoretical yield is 5.07 grams NO

The percent yield of nitrogen monoxide is 79.7 %

Explanation:

Step 1: Data given

Mass of oxygen gas = 6.74 grams

Molar mass O2 = 32.0 g/mol

Ammonia is in excess

The reaction yields 4.04 grams of nitrogen monoxide

Molar mass of nitrogen monoxide = 30.01 g/mol

Step 2: The balanced equation

4NH3 + 5O2 → 4NO + 6H2O

Step 3: Calculate moles O2

Moles O2 = mass O2 / molar mass O2

Moles O2 = 6.74 grams / 32.0 g/mol

Moles O2 = 0.211 moles

Step 4: Calculate moles NO

For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.211 moles O2 we'll have 4/5 * 0.211 = 0.1688 moles NO

Step 5: Calculate mass NO

Mass NO = 0.1688 moles * 30.01 g/mol

Mass NO = 5.07 grams

Step 6: Calculate the percent yield

Percent yield = (actual yield / theoretical yield) * 100 %

Percent yield = (4.04 grams / 5.07 grams ) * 100 %

Percent yield = 79.7 %

The theoretical yield is 5.07 grams NO

The percent yield of nitrogen monoxide is 79.7 %

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explanation:

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.