I have a question that gives two concentrations and asks for the mass of HCl formed by the reaction.

I have two concentrations:

250mL of 4.00M $\ce{H2SO4}$, and

250mL of 1.00M $\ce{NaCl}$.

Here is the balanced reaction equation:

$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCL}$

I know how to find the mass once I find the moles, molecular weights and then grams by multiplying the two.

However, how do I add those two concentrations? I assume the mixture will be at least 500mL, but how do I add the molarity?



Your working equation is correct. $$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

  • Find the amount of substance of protons of each solution via $n=c\cdot V$.

$n(\ce{H+})=2~\mathrm{mol}$, $n(\ce{Cl-})=0.25~\mathrm{mol}$

  • What is the limiting agent?


  • How many moles of hydrogen chloride can only be formed?


  • Calculate the mass of hydrogen chloride via $m = n\cdot M$

$M(\ce{HCl})=36.5~\mathrm{g/mol}$, $m(\ce{HCl})=9.1~\mathrm{g}$

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  • $\begingroup$ Martin, you are a formatting genius. $\endgroup$ – Dissenter Jun 26 '14 at 16:04
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    $\begingroup$ @Dissenter As a child I loved playing in a sandbox... so I enjoy that now and again too: Check this out! $\endgroup$ – Martin - マーチン Jun 27 '14 at 2:09

You're right; you can't directly add the two concentrations since these concentrations are specific to a certain volume of solvent.

If you want to recalculate concentrations of each system component in a combined solution, you need to find the moles of each system component in this combined solution, and then divide by the total solution volume.

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  • $\begingroup$ So does that become 500mL of 2.500 M solution? $\endgroup$ – Mark Jun 26 '14 at 3:50
  • $\begingroup$ That is not quite correct, as you have a limiting agent present. $\endgroup$ – Martin - マーチン Jun 26 '14 at 6:16

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