[imakuneee]

I have a couple of problems on our review sheet that I honestly haven't a single clue as to how to solve.

What volume of 0.0125 M HBr is required to titrate 250ml of 0.0100M Ca(OH)₂? Write the equation.

-I have the equation so far I think.
     2 HBr + Ca(OH)₂ -> CaBr₂ + 2 H₂O

Concentrated sulfuric acid with a density of 1.64 g/cm³ contains 97.5% H₂SO₄ by mass. Determine how many liters of this acid is required to manufacture 6500 Kg of fertilizer Ca(H₂PO₄)₂ by this reaction.
 2 Ca₅(PO₄)₃F + 7 H₂SO₄ + 14 H₂O -> 3 Ca(H₂PO₄)₂ + 7 CaSO₄ + 2 H₂O + 7 HF

-Well, I don't know how to solve it but I've managed to convert it into mol/L, if that's even how I'm supposed to start.

Thank you for reading and replying! 

[sjb]

What volume of 0.0125 M HBr is required to titrate 250ml of 0.0100M Ca(OH)₂? Write the equation.

-I have the equation so far I think.
     2 HBr + Ca(OH)₂ -> CaBr₂ + 2 H₂O

OK, so how much calcium hydroxide have you added? Which means how much HBr do you need? Given that you have a concentration, and an amount, you can calculate a volume.

Concentrated sulfuric acid with a density of 1.64 g/cm³ contains 97.5% H₂SO₄ by mass. Determine how many liters of this acid is required to manufacture 6500 Kg of fertilizer Ca(H₂PO₄)₂ by this reaction.
 2 Ca₅(PO₄)₃F + 7 H₂SO₄ + 14 H₂O -> 3 Ca(H₂PO₄)₂ + 7 CaSO₄ + 2 H₂O + 7 HF

-Well, I don't know how to solve it but I've managed to convert it into mol/L, if that's even how I'm supposed to start.

6500 kg of fertiliser is how many moles? Which you need how many moles of acid to make?

If you've managed to calculate the molarity of the acid, you have an amount and a concentration so you can calculate a volume.

S

[imakuneee]

Thank you very much sjb. Once I read your reply it all become very, very obvious to me. But just to make sure I solved it right, I'll post my work.

1.

.250 L * .0100 mol Ca(OH)₂/ 1 L * 2 mol HBr/ 1 mol Ca(OH)₂ * 1 L / .0125 mol HBr = 0.400 L HBr

2.

6.5 * 10⁶g Ca(H₂PO₄)₂  * 1 mol/ 234.06 g * 7 mol H₂SO₄/ 3 mol Ca(H₂PO₄)₂ = 64798.19989 mol H₂SO₄ (no sig. figs)

1.64 g H₂SO₄ / 1 cm³ * 97.5 g / 100 g * 10³ cm³ / 1 L * 1 mol H₂SO₄ / 98.09 g H₂SO₄ = 1 L / 17.1 mol H₂SO₄

1 L / 17.1 mol H₂SO₄ * 64798.19989 mol H₂SO₄ = 3.98 * 10³ L (3 sf)

Thank you again for the help

[Astrokel]

Your final answers are correct. Good luck to your exam!