elp Verify your Apple ID, davis.g.scot, 207 @gmail.com.. Durhan Technical Commun

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elp Verify your Apple ID, davis.g.scot, 207 @gmail.com.. Durhan Technical Communty Coliege CM saplinglearning.com in HD Jump to 2/14/2018 12:00 PM 0.3/s2/12/2018 02:07 PM Gra -Print al ulator Periodic Table Question 10 of 16 Sapling Learning The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many moden chemical processes. In the first step of the Ostwald process, ammonia is reacted with axygen gas to produce nitric oxide and water What is the maximum mass of HO that can be produced by combining 57.4 g of each reactant? 4NH,(g)+50 Number g H,O Hint

Explanation / Answer

1)

Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol


mass(NH3)= 57.4 g

use:
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(57.4 g)/(17.034 g/mol)
= 3.37 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 57.4 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(57.4 g)/(32 g/mol)
= 1.794 mol
Balanced chemical equation is:
4 NH3 + 5 O2 ---> 6 H2O + 4 NO


4 mol of NH3 reacts with 5 mol of O2
for 3.37 mol of NH3, 4.212 mol of O2 is required
But we have 1.794 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

According to balanced equation
mol of H2O formed = (6/5)* moles of O2
= (6/5)*1.794
= 2.152 mol


use:
mass of H2O = number of mol * molar mass
= 2.152*18.02
= 38.8 g
Answer: 38.8 g

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