What is the limiting reactant if 10 moles of NH3 react with 30.0 moles of NO? 4NH3+6NO → 5N2 + 6H2O

Explanation:

M

r

N

H

3

=

17

;

M

r

O

2

=

32

;

M

r

N

O

=

26

;

M

r

H

2

O

=

18

Using the equation:

Number of Moles = Mass / Mr

Number of Moles

N

H

3

=

175

17

=

10.3

Number of Moles

O

2

=

310

32

=

9.69

O

2

is the limiting reagent because the least moles of this are used in the reaction.

From the balanced equation, we can see that

O

2

and

N

O

are in a 5:4 ratio

5

4

=

0.8

9.69

×

0.8

=

7.75

7.75 moles of

N

O

is the maximum that can be produced from 310g of

O

2

. To convert this to mass we use the same equation as in the first step.

Mass = Moles x Mr

7.75

×

26

=

202

g

202g is the theoretical yield of

N

O

To calculate the percentage yield, you just divide the actual yield by the theoretical yield and multiply by 100:

197

202

×

100

=

98

%

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