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Topic: Combustion and Neutralization Question  (Read 3843 times)

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Offline Skychom

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Combustion and Neutralization Question
« on: April 15, 2015, 09:09:20 AM »
Poorly sorted energy waste contained 0.62 % of chlorine. 1.0.106
kg of this waste is burned.
Assume that all chlorine reacts and forms hydrogen chloride (HCl) vapor when the waste is
burned.
a) Calculate the mass of hydrogen chloride vapor formed.
b) Hydrogen chloride is removed from flue gas by scrubbing and neutralizing it with
calcium oxide CaO. Calculate the theoretical mass of CaO needed for the neutralization.
Atomic masses: MCa = 40.1, MO = 16.0, MH= 1.0, MCl = 35.5

Solutions would be greatly appreciated.

Offline Borek

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Re: Combustion and Neutralization Question
« Reply #1 on: April 15, 2015, 09:17:38 AM »
You have to show your attempts at solving the question to receive help. This is a forum policy.

Please read the forum rules.
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Offline Skychom

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Re: Combustion and Neutralization Question
« Reply #2 on: April 15, 2015, 09:51:54 AM »
My apologies.

For the first part, I was thinking that the empirical formula would be needed in order to develop the ratio of reactants:products. So we would have something like CxHyCl + O2 --> CO2 + H2O + HCL

I cannot seem to know how to solve for the unknowns here.

For the second part, I believe we are looking at the reaction: 2 HCl + CaO --> CaCl2 + H2O

Since it is a 2mol:1mol ratio, one could use the weight of HCl from a) to solve for CaCl2 by doubling it?
« Last Edit: April 15, 2015, 10:03:20 AM by Skychom »

Offline Borek

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Re: Combustion and Neutralization Question
« Reply #3 on: April 15, 2015, 10:36:17 AM »
All you need is just a mass (and number of moles of chlorine), for that you don't need any reaction equation. Cl :rarrow: HCl, molar ratio is obvious.

For b - yes, that's the correct reaction equation, though I have no idea no idea what you mean by

solve for CaCl2 by doubling it?
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Offline Skychom

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Re: Combustion and Neutralization Question
« Reply #4 on: April 18, 2015, 12:00:10 PM »
Thank you for your reply.

I am still uncertain about part a) since I am not given the molar mass of the 'energy waste' reactant. You seem to be implying that it is not needed, but I am hoping for some clarification. Typically I would solve this problem by determining the moles of the reactant (n = m / M) and use the 1:1 mole ratio of Cl -> HCl to then figure out the mass of HCl so that:

(n of waste = n of HCl) m of HCl = n * M

For part b), am I correct in stating that the mass of CaO would by: m = (nHcl*2) * M of CaO ?

Offline Borek

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Re: Combustion and Neutralization Question
« Reply #5 on: April 20, 2015, 11:28:31 AM »
What does 0.62% mean? How many kilograms of chlorine in the waste?

Once you know the mass of chlorine, you can easily convert it to moles.
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Offline Skychom

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Re: Combustion and Neutralization Question
« Reply #6 on: April 20, 2015, 02:52:19 PM »
0.62% would mean 0.62g of Cl in 100mL of waste, so n = 0.62g / 35.5 = 0.0175 mol

So for b) n CaO = 2*nCl = 0.035

m = n*M = 0.035*56.1 = 1.96g ?

Thanks again

Offline Borek

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Re: Combustion and Neutralization Question
« Reply #7 on: April 20, 2015, 03:53:31 PM »
You said you were given mass of the waste (although what you wrote - 1.0.106 kg - doesn't make much sense as a number edit: was it 1.0×106 kg?). Why do you use a volume of 100 mL?
« Last Edit: April 20, 2015, 04:22:00 PM by Borek »
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