Ask Me Help Desk - Sufficient reagent

How can I show that 6ml of 6.0M HCl would or would not be sufficient to react with 3.8gNaHCO3 in

NaHCO3 + HCl---yields-----NaCl + H2O + CO2

1. Figure out the number of moles of NaHCO3

M=mass of NaHCO3 = 3.8 g
MW = Molecular Mass (Molecular Weight) of NaHCO3 (in grams/mole)

$\cancel {grams} \,\div\,\ \frac {\cancel {grams}}{mole} = moles\,of\, NaHCO_3$

Multiply the number of moles of NaHCO3 by 2 (from your balanced equation) because you need 1 mole of HCl for 1 mole of NaHCO3. This is the number of moles of HCl that are required.

$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO2$

$Moles\,of\,HCl\,required = 1\,\times\, moles\,of\,NaHCO_3$

2. Figure out how many moles of HCl you have.

$Moles\,of\,HCl = \frac {6 \, Moles}{\cancel {Liter}} \,\times\, 6\,\cancel {mL} \,\times\, \frac {1\,\cancel {Liter}}{1000\,\cancel {mL}} =$

This is the number of moles of HCl that you actually have. You can now compare this number with the number of moles that are required to get your answer.

ankara55t, could you please stop creating threads involving the same question? It's difficult to follow, and you happen to have many 'linked' questions, and some are even duplicates. For example, group all those concerning sodium hydrogen carbonate (NaHCO3) and hydrochloric acid (HCl) in a single thread, and those concerning potassium carbonate (K2CO3) and hydrochloric acid in another.

Thanks!

Yes. Sorry I will stop doing that. I am grateful to Perito for his help.