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1- How many milliliters of 0.140 M HCl are needed to completely neutralize 42.0mL of 0.106 M Ba(OH)2 solution?

2-How many milliliters of 0.123 M H2SO4 are needed to neutralize 0.250 g of NaOH ?

3-If 55.2 mL of BaCl2 solution is needed to precipitate all the sulfate ion in a 750 mg sample of Na2SO4 , what is the molarity of the solution?

4-If 42.7 mL of 0.218 M HCl solution is needed to neutralize a solution of Ca(OH)2 , how many grams of Ca(OH)2 must be in the solution?

1- How many milliliters of 0.140 M HCl are needed to completely neutralize 42.0mL of 0.106 M Ba(OH)2 solution?

Step 1. Write a balanced equation.
2HCl + Ba(OH)2 ==> BaCl2 + 2H2O

Step 2. How many moles of the base do we have. moles = M x L = 0.106 M x 0.042 L = 0.00445.

Step 3. How many moles of HCl are needed? Use the coefficients in the balanced equation to do this. We have 0.00445 moles Ba(OH)2. It takes twice as much HCl as it does Ba(OH)2; therefore, we must have used 2 x 0.00445 moles HCl = 0.00890 mols HCl.

Step 4. Remember the definition of molarity. M = moles/L. For the HCl we have 0.00890 moles and the molarity is 0.140 M, so solve for L.
L = mols/M = 0.00890/0.140 = 0.0636 L or 63.6 mL.

All of the solution problems are worked the same way. If the problem throws in a solid instead of a solution, just remember that moles = grams/molar mass. After converting into moles, everything is worked the same way. Try it, you'll like it.

On step 3 you divide by 2 not multiply.

How many ml of 9 M HCl are needed to neutralize 0.500 L of 0.350 M NaOH? Given the reaction CH4+2O2=C02+2H2O

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