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

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: