If 1.00 g of NH_3 and 1.50 g of O_2 are mixed, which is the limiting reactant? 4NH_3(g) + 5O_2(g) rightarrow 4NO(g) + 6H_2O(s) (b) What is the theoretical yield (in grams) of NO that can be produced when the quantities in part a are mixed? (c) If 1.05 g of NO are actually obtained from the reaction, what is the percent yield? Find number of N_2O molecules and mass of N in 1.50 g of N_2O sample.
4NH3(g) + 5o2(g) —-> 4NO(g) + 6H2O(s)
4 mole NH3 = 5 mole O2
No of mole of NH3 = 1/17 = 0.0588 mole
No of mole of O2 = 1.5/32 = 0.047 mole
a) limiting reactant = O2
b) theoretical yield of NO = 0.047*4/5 = 0.0376 mole
= 0.0376*30 = 1.128 g
c) percent yield = (practical yield/theoretical yield)*100
= 1.05/1.128*100
= 93.1%
6)
No of mole of N2O = 1.5/44 = 0.034 mole
No of molecules of N2O present = 0.034*6.023*10^23
= 2.048*10^22
mass of N = 1.5*28/44 = 0.9545 g
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