In the equation: 4NH3 + 5O2 → 4NO + 6H2O  , 310 g of O2 will react with 175 g of NH3. What is the limiting reactant? What is the theoretical yield of NO, and if 197 g of NO are produced, what is the percent yield?

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Answer 1 · 2018-01-09T11:26:59

Theoretical yield = 202g; Percentage yield = 98%

#Mr NH_3 = 17; Mr O_2 = 32; Mr NO = 26; Mr H_2O = 18#

Using the equation:

Number of Moles = Mass / Mr

Number of Moles #NH_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 #NO# are in a 5:4 ratio

#5/4 = 0.8#

#9.69 xx 0.8 = 7.75#

7.75 moles of #NO# 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 xx 26 = 202g#

202g is the theoretical yield of #NO#

To calculate the percentage yield, you just divide the actual yield by the theoretical yield and multiply by 100:
#197 / 202 xx 100 = 98%#