Answer:
Alright the very first thing you need to do is balance the equation:
2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O
Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.
Volume x Concentration/molarity = moles
0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl
0.094L x 0.75 M = 0.0705 moles /molar ratio of 1 = 0.0705 moles of Na2CO3
Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.
Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.
0.235 + 0.094 = 0.329L of NaCl
Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.
(0.75M) x (0.094L) = C2 x (0.329L)
Rearrange equation to solve for C2:
<u>(0.75M) x (0.094L)</u> = C2
(0.329L)
C2 = 0.214 M (Rounded)
<u>When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.</u>
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