In the reaction: NH3 + O2 → NO + H2O How many grams of NO are produced?

Question

Consider the reaction: NH3 + O2 → NO + H2O

A) How many grams of NO are produced from the complete reaction of 160.00 g of O2 ?

B) If the percent yield of this above reaction is found to be 92.98 %, how many grams of NO are actually produced ?

...thank you very much for your time :)

Answer 1

The first thing you need is a balanced equation for the reaction:

4 NH3 + 5 O2 → 4 NO + 6 H2O

A) 5 moles of O2 produce 4 moles of NO

160.00 g of O2 / g/mol for O2 = number of moles of O2

160.00 g / 16.00 g/mol = 10.0 moles of O2

This would produce 8.0 moles of NO

8.0 moles * 30.01 g/mol = 240.08 grams of Nitric oxide.

B) If the percent yield is only 92.98 % then

240.08 g * 0.9298 = 223.226 grams (you might want to round this to 223.2 grams).

Answer 2

First responder has a powerful factor. yet i will nevertheless define the attitude to the respond. The protocol for all problems of this sort is often the comparable: first see if the reaction is balanced (it is not), and whether it is not, restoration it. next compute the molecular weights of the significant species (here, H2O, and in spite of it fairly is that weighs sixty 8.12 grams). Then compute the style of moles of reactant, which provides you with the style of moles of product, and multiply that by using the suitable molecular weight to end the interest.