Question

Consider the following UNBALANCED equation: NO(g) + H2(g) → NH3(g) + H2O(g) If you start with 89.3 g NO(g) and 28.6 g H2(g), find the theoretical yield of ammonia. Express your answer in grams. 8. (2 Points) Ron Weasley isn’t the best experimentalist. After an experiment he ends up with only 3.10 g of product. However, the same reaction has is a theoretical yield of 3.50 grams of product. What is the percent yield?

Answer 1

Answer:

50.74 g NH3 and  88.6 %

Explanation:

This is a problem based on stoichiometric calculations.

First we will need the balanced equation of the reaction, then calculate the moles of the reactants and determine the limiting reagent if any. Finally we will calculate the theoretical yield. For the second part we will just calculate the percent yield form the quantities given.

           2 NO + 5 H2 ======== 2  NH3 + 2 H2O

89.3 g x  1 mol NO/ 30.0 g  = 2.98 mol NO  (MW NO: 30.0 g/mol)

28.6 g  x 1 mol H2/ 2.0 g  = 7.45 mol H2     ( MW H2 :   2.0 g/mol)

The limiting reagent is:

5 mol H2 required/ mol NO = 7.45 mol H2

we have almost twice that amount therefore NO is the limiting reagent

mol NH3 produced will be:

2.98 mol NO x 2 mol NH3/ mol NO = 2.98 mol NH3

and its mass is

17.03 g NH3 mol NH3 x 2.98 mol NH3 = 50.74 g NH3

( MW NH3 : 17.03 g/mol)

For the second part,

% yield = 3.10g/3.50 x 100 = 88.6