calculating masses...grade 11 Chem...?

Question

1. The Solvay process for the manufacture of sodium carbonate involves the following reactions:
NH3(g) + H2O(l) + CO2(g) -(arrow sign) NH4HCO3(aq)
NH4HCO3(aq) + NaCl(aq) -(arrow sign) NH4Cl(aq) + NaHCO3(s)
2NaHCO3(s) -(arrow sign) Na2CO3(s) + H2O(g) + CO2(g)

Calculate the masses of ammonia and sodium chloride required to produce 150.0kg of pure anhydrous sodium carbonate.

Note:I have been stuck on this one question for 2 hrs, 15 minutes and I just can't work it out correctly. I have a large chemistry text books and another booklet. I don't know the processes involved in finding the answers. Could you please show me the correct working and answers and then at the bottom explaining it all to me. I'm NOT trying to get you to do my homework. I just need serious help! Please help chem whizzes! Thankyou so much. Best answer gets 10 points!

Um...it's only 4:49pm where I live so yeah..I'm not going to go to bed yet.

I am receiving different answers from most of you. Could you please tell me which one is correct (please be 100% sure). Thankyou.

Answer 1

Molar mass of Na2CO3 = 2*23 + 12 + 16*3 = 106 g

Molar mass of NaCl = 23 + 35.5 g = 58.5 g

Molar Mass of NH3 = 14 + 3 = 17 g

Now 1 mol of Na2CO3 needs 2 mol of NH3 and 2 mol of NaCl

(Multiply eqn, 1 and 2 by 2 and add to 3)

Thus 150 kg is eqaul to 150*1000 / 106 mol = 1415.09 mol

That means 2830.18 mol each of NaCl and NH3 are required.

That means 2830.18 x 58.5 g = 165.57 kg of NaCl

and 2830.18 x 17 g = 48.11 kg of NH3

Answer 2

the last three answers are wrong the second answer is the only one is correct


use the Hess law method to find the overall equation
which is:

NaHCO(s)+NH3(g)+NaCl(aq)
+H2O(l)-->

NH4Cl(aq)+Na2CO3(s)+H2O(g)


So, base on this information one mole of NaCO3 reacts with NaCl and NH3 on an 1:1 ratio factor.

just convert moles to kg


150kg Na2CO3
(1molNa2CO3/.106kgNa2CO3)=

1415.1mol Na2CO3


so mass of NaCl needed is:

.05845kg NaCl*1415.1mol =87.71kg NaCl

and for NH3 is .017kg*1415.1mol = 24.1kg NH3

but my anwser is 100% correct

Answer 3

NH3 + H2O + CO2 -> NH4HCO3
NH4HCO3 + NaCl -> NH4Cl + NaHCO3
2NaHCO3 -> Na2CO3 + H2O + CO2

By adding all the reaction..

2NH3 + 2H2O + 2CO2 + 2NaCl -> 2NH4Cl + Na2CO3 + H2O + CO2

Nah.. Let see..
mole of sodium carbonate produced = 150000 / 106 = 1415.1 moles
mole of NaCl and NH3 needed = 2 * 1415.1 = 2830.2 moles

By converting the moles to mass,
mass of NH3 needed = 2830.2 * 17 = 48,1 kg
mass of NaCl needed = 21830.2 * 58.5 = 165,5 kg


peace
vixklen

Answer 4

Last two answers are correct.
2 moles NH3..........................................1 mole Na2CO3
2 moles NaCl.......................................... 1 mole Na2CO3
mass NH3 = (2*17*150/106)= 48.1132g
mass NaCl =(2*58.5*150/106)=165.56603g

Where: mass of x = [2* (M of x )* (m of Na2CO3)/ (M of Na2CO3)]

PS: Need to put = sign for equilibrium!

Answer 5

it is very simple

molecular mass of NH3 is 17
molecular mass of NaCl is 58.5
n
molecular mass of Na2CO3 is 106

we will use unitary method

if u see d reactions u will see that 1 molecule of NH3 n 1 molecule of NaCl is needed to make 1 molecule of Na2CO3.......i.e 17 kg of NH3 n 58.5 kg of NaCl to make 106 kg of Na2CO3....


therefore 106kg == 17 kg of NH3
1 kg == 17/106 kg of NH3
150 kg of Na2CO3 == (17/106) * 150 kg of NH3
==24.06 kg of NH3

same way 150 kg == (58.5/106) * 150 kg of NaCl
== 82.78 kg on NaCl

Answer 6

go to bed