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## Main Question or Discussion Point

What volume of .273 Molarity HCl is required to react completely with 50.00 mL of .167 Molarity of Barium Hydroxide?

2 HCl + Ba(OH)2 -> 2 H20 + BaCl

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I am confused about this problem because my profe told me that you need to double the amount of Hydrogen b/c there are 2 moles of HCl. Can someone explain this?

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Also, my profe solved the problem by this method:

50.00 mL x L/1000 x .167 mol = 8.35 x 10^-3 mol Ba(OH)2 which means Ba is 8.35 x 10^-3 mol. OH is 8.35 x 10^-3 mol.

This is where I get confused.

Then he took 2 mol HCl x (8.35 x 10^-3 OH) x (1L/.273) mol x 10^3 mL of HCl.

I don't understand why he used HCl with OH and how he canceled the OH out to get the HCl.

Thanks for the help!

2 HCl + Ba(OH)2 -> 2 H20 + BaCl

--------------------------------------------

I am confused about this problem because my profe told me that you need to double the amount of Hydrogen b/c there are 2 moles of HCl. Can someone explain this?

----------------------------------------

Also, my profe solved the problem by this method:

50.00 mL x L/1000 x .167 mol = 8.35 x 10^-3 mol Ba(OH)2 which means Ba is 8.35 x 10^-3 mol. OH is 8.35 x 10^-3 mol.

This is where I get confused.

Then he took 2 mol HCl x (8.35 x 10^-3 OH) x (1L/.273) mol x 10^3 mL of HCl.

I don't understand why he used HCl with OH and how he canceled the OH out to get the HCl.

Thanks for the help!