It is desired to produce 4.74 grams of nitrogen dioxide by the following reaction. If the percent yield of nitrogen dioxide is 76.0 %, how many grams of nitrogen monoxide would need to be reacted? grams nitrogen monoxide nitrogen monoxide (g) + oxygen (g) nitrogen dioxide (g)

We need 4.08 grams of NO to be reacted

Explanation:

Step 1: The balanced equation

2NO (g) + O2 (g) →2NO2 (g)

Step 2: Data Given

Mass of NO2 produced = 4.74 grams = actual yield

percent yield = 76.0 %

Molar mass of NO = 30.01 g/mol

Molar mass O2 = 32 g/mol

Molar mass NO2 = 46 g/mol

Step 3: Calculate theoretical mass of NO2

Yield = 0.76 = actual % / theoretical %

Theoretical % = 4.74 / 0.76 = 6.24 grams

Step 4: Calculate moles of NO2

Moles of NO2 = mass NO2 / Molar mass NO2

moles of NO2 = 6.24 grams / 46 g/Mol

moles of NO2 = 0.136 moles

Step 5: Calculate moles of NO

For 2 moles of NO to be consumed, we have 2 moles of NO2 to be produced.

So we have 0.136 moles of NO2 produced, and 0.136 moles of NO consumed

Step 6: Calculate mass of NO

mass = Moles NO * Molar mass NO

mass NO = 0.136 mol * 30.01 g/mol = 4.08 grams

We need 4.08 grams of NO to be reacted