Help me.help me. Chemistry questions.i gonna be mad now.?

Question

How much water must be added to 294 grams of H2SO4 to form a 6 molar solution of Sulfuric Acid

500 mL of 9 M Sulfuric Acid (H2SO4) is combined with 750 mL of 6 M Barium Hydroxide (Ba(OH)2 producing Barium Sulfate (BaSO4) and water. What mass of Barium Sulfate will be produced. (assume 100% yield)

How does the H+ ion concentration of a solution that has a ph of 3 compare with the H+ ion concentration of a solution with a ph of 6.

Answer 1

Moles H2SO4 = 284 g / 98 g/mol =2.90

M = mol/V

V = mol/M = 2.90 / 6 = 0.483 L => 483 mL

Moles H2SO4 = 0.500 L x 9 =4.5

Moles Ba(OH)2 = 0.750 L x 6 =4.5

H2SO4 + Ba(OH)2 >> BaSO4 + 2 H2O

the ratio is 1 : 1

moles BaSO4 = 4.5

mass BaSO4 = 4.5 mol x 233.43 g/mol =1050 g

at pH = 3 => [H+] = 1.0 x 10^-3 M

at pH = 6 => [H+] = 1.0 x 10^-6 M

Answer 2

1.  A 6-M solution contains 6 moles per liter (by definition of "molar").  How many moles sulfuric acid do you have?  You are given the mass sulfuric acid, do you remember how to convert?

2.  750 mL = 0.75 liter

6 moles/liter * .75 liter = 4.5 moles barium hydroxide

Do the same for sulfuric acid

Determine the limiting reagent.

How many moles of barium sulfate can you produce with these amounts of reagents?

How much does that weigh?

3.  Remember that pH = - log [H+]

DO NOT DO THE WORK FOR THE STUDENT.