how many mL of .135M Sr(OH)2 are needed to neutralize 35ml of .157M HCL
I know the answer came out to 20.4 ml
My only question is how do I convert HCL to Sr(OH)2
You've asked two or three questions about how to make the conversions. They're all done with dimensional factors.
Write and balance the equation.
Sr(OH)2 + 2HCl ==> SrCl2 + 2H2O
This equation TELLS you that 1 mol Sr(OH)2 reacts with 2 mols HCl to form 1 mol SrCl2 and 2 mol H2O. So those factors are used to make the conversion. Knowing any one of the four you can calculate all of the others.
mols HCl = M x L = 0.157 x 0.0350.005495
0.005495 mol HCl x (1 mol Sr(OH)2/2 mol HCl) = 0.005495 x 1/2 = 0.0027475
Then M Sr(OH)2 = mols/L
mols = 0.0027475; L = ?; M = 0.135
So L = 0.0027475/0.135 = 0.02035L = 20.35 mL. You are allowed 3 significant figures so I would round this to 20.4 mL.
Hope this helps. Knowing mols HCl allows you to use factors to calculate mols of anything else in the equation.
why is it 2HCL
The equation must be balanced.
To convert the amount of one substance to another, we need to use stoichiometry, which is based on the balanced equation of the chemical reaction.
The balanced equation for the reaction between HCl (hydrochloric acid) and Sr(OH)2 (strontium hydroxide) is as follows:
2 HCl + Sr(OH)2 -> SrCl2 + 2 H2O
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Sr(OH)2. This means the stoichiometric ratio between HCl and Sr(OH)2 is 2:1.
Given that the volume of HCl is 35 mL and the concentration is 0.157 M, we first need to calculate the number of moles of HCl present.
Number of moles of HCl = Volume (in liters) x Concentration
Number of moles of HCl = 35 mL x (1 L / 1000 mL) x 0.157 M
Number of moles of HCl = 0.005495 moles
Since the stoichiometric ratio is 2:1 (HCl:Sr(OH)2), we can calculate the number of moles of Sr(OH)2 needed.
Number of moles of Sr(OH)2 = 0.005495 moles / 2
Number of moles of Sr(OH)2 = 0.002748 moles
Now, we can find the volume of Sr(OH)2 needed using the concentration of Sr(OH)2.
Volume of Sr(OH)2 = Number of moles x Concentration (in liters)
Volume of Sr(OH)2 = 0.002748 moles x (1 L / 1000 mL) x 0.135 M
Volume of Sr(OH)2 = 0.000371 mL
So, you would need approximately 0.000371 mL of 0.135 M Sr(OH)2 to neutralize 35 mL of 0.157 M HCl.
Since the volume of Sr(OH)2 is very small compared to the volume of HCl, it is recommended to check the calculations and ensure that the values entered for the concentrations and volumes are accurate.