- #1

Raza

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**1.What volume of a 15.0mol/L standard solution of hydrochloric acid will be required a make a 450mL of a 2.25mol/L dilution?**

[tex]\frac{15mol/L}{2.25mol/L}=6.67[/tex] *I think there is no unit for 6.67

[tex]=(6.67)(0.450L)=3L[/tex]

**2.Will it be possible for you to produce 850mL of a 1.85mol/L acetic acid sloution if all you left in your orginal 7.50mol/L standard solution bottle is 200mL?**

These long questions just confuse me. I don't know where to start on this question.

**3.Consider the following acid-base reaction:**

H

_{2}SO

_{4(aq)}+ Mg(OH)

_{2(aq)}-------> MgSO

_{4(aq)}+ 2H

_{2}O

_{(l)}

**a) Determine how many grams of magnesium hydroxide you would need to add to water in order to obtain 5.50L of a 2.25mol/L magnesium hydroxide solution.**

[tex]\frac{5.50L}{2.25mol/L}=2.44mol[/tex] *Molar mass for Mg(OH)

_{2(aq)}is 58.31968g

[tex]2.44mol[/tex] x [tex] (\frac{58.32g}{1mol})=142.3g[/tex]

**b)Suppose you had 6.00L of a 3.15mol/L sulfuric acid solution and 4.50L of a 7.00mol/L magnesium hydroxide solution. Which one of these would be the limiting reangent? How much of the excess reangent will be left unreacted?**

[tex]\frac{6.00L}{3.15mol/L}=1.94mol [/tex] - Sulfuric Acid Solution

[tex]\frac{4.50L}{7.00mol/L}=0.64mol [/tex] - Magnesium Hydroxide Solution

I say the limiting reangent is the magnesium hydroxide solution and 1.3mol of sulfuric acid solution will remain unreacted.

**c)Using the volumes and concentrations for part (b), how many grams of water will the reaction produce?**

1.3mol *Molar mass ofH

_{2}SO

_{4(aq)}is 98.03g

[tex]1.30mol[/tex] x [tex] (\frac{98.03g}{1mol})=127.44g[/tex]

Please please check my work. It is worth a lot of marks.

Thank You