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moles and balanced equation homework (AS)

I did some homework, but the tutor didn't look it. It was mainly the moles/concentration/volume bit I wasn't sure on, so I thought I'd ask here to see if I got it right or not. Also, may as well throw in my balanced equation. Did I get the state symbols right too?

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Write a balanced chemical equation for the reaction between sodium carbonate and hydrochloric acids. Show the state symbols.

Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)

What threw me off was whether to include water with the reactants, since sodium carbonate is mixed with distilled water first.

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The moles bit, at first I think I overcomplicated it by trying to work out a concentration within a concentration, so then I redid it in a simpler, more exam-y method with the old belief that if the working out is complicated, I'm doing it wrong. So:

Calculate the moles of 23.8cm^3 sodium carbonate at concentration 0.15mol/dm^3

moles (n) = concentration (c) x volume (v) = 0.15 x (23.8/1000) = 0.850mol (to 3 S.F.)

Using this calculation and your balanced equation, calculate the number of moles of hydrochloric acid that were required to neutralize the sodium carbonate solution.

Ratio of Na2CO3 to HCL is 1:2, so moles of hydrochloric acid = 0.850 x 2 = 1.7mol

Based on this, calculate the concentration of the 25cm^3 of hydrochloric acid solution in mol/dm^3.

c = n/v = 1.7/(25/1000) = 68mol/dm^3

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Did I do good?

Reply 1

BJack

Original post by ozzyoscy

xYour balanced symbol equation is fine. The water that the HCl and sodium carbonate are dissolved in don't need to be included in the equation — it's not reacting, after all.

You've taken the right approach for the moles calculation, but your answer is completely wrong. I don't know what you've done. Try recalculating it and see if you get the same answer. You might have realized something was up when you got such a high value for the concentration of the HCl. You know the ratio is 1:2 and you've got about the same volume of HCl solution as sodium carbonate solution, so your final answer should be somewhere around 0.3 M.

Reply 2

ozzyoscy

Original post by BJack

You've taken the right approach for the moles calculation, but your answer is completely wrong. I don't know what you've done. Try recalculating it and see if you get the same answer. You might have realized something was up when you got such a high value for the concentration of the HCl. You know the ratio is 1:2 and you've got about the same volume of HCl solution as sodium carbonate solution, so your final answer should be somewhere around 0.3 M.