I am having trouble finding how much excess reactant remains after the limiting

Question

I am having trouble finding how much excess reactant remains after the limiting reactant is consumed (part c). Step by step would be amazing. Thanks!

One of the steps in the commercial process for converting ammonia to nitric acid involves the conversion of NH3 to NO, as showrn below. 4 NH3(g) + 52(g) 4 NO(g) + 6 H2O(g) In a certain experiment, 3.90 g of NH3 reacts with 4.45 g of O2 (a) Which reactant is the limiting reactant? NH3 02 (b) How many grams of NO form? 333any grams of No form? 3.339 3.33 9 (c) How much of the excess reactant remains after the limiting reactant is completely consumed?

Explanation / Answer

The reaction is 4NH3+5O2----> 4NO + 6H2O

atomic weights = N :14, H=1 and O=16

Molecular weights : NH3=14+3=17, O2=32 NO= 12+16=28 and water =6*18=108

Molar ratio of reactants : 4 :5 = 1:1.25

molar ratio supplied   3/17 : 4.45/32 =0.1765 : 0.139 = 0.1765/0.1765 : 0.139/0.1765 =1 :0.78

rquired ratio is 1: 1.25 and supplied ratio is 1:0.78

hence oxygen is suppled in less amount and is the limting reactant

5*32 gms of oxygen gives rise 4*28 gms of NO

4.45 gm of oxygen gives 4*28*4.45/(5*32)=3.115 gm

160 gms of oxygen requires 4*17=68 gms of NH3

4.45 gms of oxygen requires 4.45*68/160=1.89 gms of of NH3

NH3 remaining=3.9-1.89= 2.01 gms

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.