Find the volume of 0.110 M hydrochloric acid necessary to react completely with 1.54 Al(OH)3.
Please explain how to do this.
To find the volume of 0.110 M hydrochloric acid necessary to react completely with 1.54 Al(OH)3, you need to use the balanced equation for the reaction and the concept of stoichiometry.
Let's start by writing the balanced chemical equation for the reaction you are referring to:
2 Al(OH)3 + 6 HCl -> 2 AlCl3 + 6 H2O
From the balanced equation, we can see that it takes 2 moles of Al(OH)3 to react with 6 moles of HCl.
Next, we need to convert the given concentration of HCl (0.110 M) into moles per liter. This can be done by using the equation:
moles = concentration (M) x volume (L)
Since we want to find the volume of HCl, we rearrange the equation:
volume (L) = moles / concentration (M)
Now, we need to determine the number of moles of HCl required to react with 1.54 moles of Al(OH)3. From the balanced equation, we know that the stoichiometric ratio between Al(OH)3 and HCl is 2:6. Therefore, the number of moles of HCl required will be:
moles of HCl = (moles of Al(OH)3 x 6) / 2
Substituting the given value of moles of Al(OH)3 (1.54), we can calculate the moles of HCl needed.
Finally, we can find the volume of HCl using the equation:
volume (L) = moles of HCl / concentration (M)
Substituting the calculated value of moles of HCl and the given concentration of HCl (0.110 M), we can solve for the volume of HCl in liters.
Just follow the steps and/or see your other post above under Brittany.
http://www.jiskha.com/science/chemistry/stoichiometry.html