Answer to Question #307615 in General Chemistry for Fish

Question

In a certain reaction, 1.5g of ammonia was given and 2.75g of oxygen, how much NO is produced? (in mass and mole)

4NH3 + 502 --> 4NO + 6H20

Answer 1

Since we're given the masses of the two reactants, we have to find the limiting reagent. The limiting reagent is one which will get exhausted first and thus determine the mass of the products.

Number of moles of NH3 = 1.5/1.7 = 0.088

Then we divide the number of moles by the coefficient of NH3 in the equation

=0.088/5 = 0.022

Number of moles of O2 = 2.75/32 = 0.86

Then we divide the number of moles by the coefficient of O2 in the equation = 0.086/5 = 0.017

The smallest number is 0.017, thus the limiting reagent.

From the equation,

5 moles of O2 = 4 moles of NO

0.086 moles of O2 = 0.0688 moles of NO

Number of moles = mass / molar mass

0.0688 = mass / (14 + 16)

Mass = 0.0688*30

mass = 2.064g