Respuesta :
Answer : The mass of [tex]NaHCO_3[/tex] needed is, 12.7 grams.
Explanation :
First we have to calculate the moles of HCl.
[tex]\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of }HCl[/tex]
[tex]\text{Moles of }HCl=6.01M\times 0.0252L=0.151mol[/tex]
Now we have to calculate the moles of [tex]NaHCO_3[/tex]
The balanced neutralization reaction will be:
[tex]HCl+NaHCO_3\rightarrow NaCl+H_2O+CO_2[/tex]
From the balanced chemical reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of [tex]NaHCO_3[/tex]
So, 0.151 mole of HCl neutralizes by 0.151 mole of [tex]NaHCO_3[/tex]
Now we have to calculate the mass of [tex]NaHCO_3[/tex]
[tex]\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3[/tex]
Molar mass of [tex]NaHCO_3[/tex] = 84 g/mole
[tex]\text{ Mass of }NaHCO_3=(0.151moles)\times (84g/mole)=12.7g[/tex]
Therefore, the mass of [tex]NaHCO_3[/tex] needed is, 12.7 grams.
