With the equation:
HCl + NaHCO3 --> NaCl + CO2 + H2O
Find the number of grams of NaHCO3, to three significant figures, required to neutralize 75 mililitres of 0.110 mol/l HCl.
I'm stuck on this problem. I think I either use the molarity formula or c=n/v?
If someone could show or give me a hint on how to start the problem I think I can figure the rest out.
Thank you in advance!
Since it's an easy question, let me venture an answer.
First find the amount in moles of $\rm{HCl}$ you have to neutralise.
In $75$ ml of the given acid solution, you will have $75 \times 10^{-3} \times 0.110 = 8.25 \times 10^{-3}$ mol of $\rm{HCl}$.
You will need exactly that many moles of $\rm{NaHCO_3}$ to neutralise that amount of acid since the stoichiometric ratio as per the balanced equation is $1:1$
The molecular "weight" (or mass) of the $\rm{NaHCO_3}$ molecule is $84$ (just add the atomic masses - $23 + 1 + 12 + 3 \times 16$. Which means $1$ mole of $\rm{NaHCO_3}$ weighs very close to $84$ gram.
$8.25 \times 10^{-3}$ mol of $\rm{NaHCO_3}$ has a mass of $8.25 \times 10^{-3} \times 84 = 0.693$ gram.