4NH3 + 5O2 = 4NO + 6H2O How many grams of O2 are needed to completely react with 3.315 moles of NH3?

Question

Grateful for any help given!

Answer 1

Hi kitty I'm more than happy to help

In this question you're given the equation because you will need it when applying the concept of stiochiometry:

n (O2) / 5 = n (NH3) / 4

If you realize 5 is the number next to the oxygen gas in the equation: this indicates that for every 5 Oxygen gas molecules you will need 4 ammonia molecules. Sometimes, they give you the equation not balanced so you have to balance it!

From here we can calculate the number of moles for the oxygen gas

By rearranging this equation: n (O2)/ 5 = n (NH3) / 4 you get the following: n (O2) = 5 n (NH3) / 4

Now all what you have to do is substitute the number of moles of NH3 in the equation and you will get the following:

n (O2) = 5 x 3.315 moles / 4 = 4.14375 moles

But the question didn't end here because they're asking about the mass and not the number of moles. So, we will use the number of moles to convert in to mass by using the molar mass of O2

Molar mass of O2 = 2 x Molar mass of Oxygen = 2 x 16g/mol = 32 g/mol

mass (O2) = n (O2) x M (O2) = 4.14375 mol x 32 g/mol = 132.6 g of O2

Answer: 132.6 g of O2

I hope the explanation helped you, Have a good day and good luck!

Answer 2

We will follow these steps

1. Make sure you have a balanced chemical equation (yes, its balanced)
2. use mole ratio to convert from moles of NH3 to moles of O2.
3. The ratio is 4:5. 4 mol NH3: 5 mol O2We get this from the coefficients from the balanced equation
4. 3.315 mol NH3 * (5mol O2/ 4mol NH3) = 4.14375 mol O2
5. Use molar mass to convert moles of O2 to grams of O2
6. molar mass of O2 is 32.00g
7. 4.14375 mol O2*(32.00g O2/1molO2) = 132.6 g O2