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We were given this problem at my university, and I can't seem to figure out how it was solved. The question says we have $\ce{NaHCO3}$ and 100 ml of $\ce{HCl}$ react and one of their products is $\ce{CO2}$ gas whose volume is 11.2 L, find the molar concentration of $\ce{HCl}$.

I first found the moles of $\ce{CO2}$ by dividing 11.2 L / 22.4 L which gave me 0.5 mol. Then I know concentration is number of moles/volume, so should I assume the moles $\ce{HCl}$ are also 0.5 by ratio and divide 0.5 by 0.1 L?

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    $\begingroup$ Try to solve it yourself first. Don't give up so easily. Edit your post showing your efforts. $\endgroup$ – Pritt Balagopal May 30 '17 at 17:52
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Note: This is a homework question, so I will not provide a complete answer.

You have found the amount of $\ce{CO2}$, where did it come from? If you guessed sodium bicarbonate, congratulations, you win a cookie! Now you should be knowing the amount of $\ce{NaHCO3}$ that underwent the reaction.

Write the balanced equation for the reaction of $\ce{HCl}$ and $\ce{NaHCO3}$. Assuming that $\ce{NaHCO3}$ is present in excess, how much $\ce{HCl}$ was present in that $\pu{100mL}$ sample. What is the concentration now?

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  • $\begingroup$ Ok, if NaHCO3 was in excess the balanced equation would be 1NaHCO3 + 0.5HCL = 0.5CO2 because the reaction ratio is still 1:1 so even if sodium bicarbonate was in excess , the HCL would be limiting how much HCL is produced $\endgroup$ – macphail magwira Jun 3 '17 at 1:11

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