3H_2(g)+N_2(g)--->2NH_3(g)?
3H2(g)+N2(g)--->2NH3(g)
Given:
1.81 g H2 is allowed to react with 9.96 g N2, producing 1.13 g NH3.
1.)What is the theoretical yield for this reaction under the given conditions?
2.)What is the percent yield for this reaction under the given conditions?
More Questions From This User See All
Answers & Comments
Verified answer
moles N2 = 9.96 g / 28.0134 g/mol=0.356
moles H2 needed = 0.356 x 3 =1.07
actual moles H2 = 1.81 / 2.016 =0.898 => limiting reactant
theoretical moles NH3 = 0.898 x 2 /3=0.599
theoretical mass NH3 = 0.599 x 17.0307 g/mol=10.2 g
% = 1.13x 100/ 10.2 =11.1