Is My Grade 11 Chemistry Homework Correct?

[Raza]

For this one, Could someone check my work?

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?

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

$$=(6.67)(0.450L)=3L$$



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:
H₂SO_4(aq) + Mg(OH)_2(aq) -------> MgSO_4(aq) + 2H₂O_(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.

$$\frac{5.50L}{2.25mol/L}=2.44mol$$ *Molar mass for Mg(OH)_2(aq) is 58.31968g

$$2.44mol$$ x $$(\frac{58.32g}{1mol})=142.3g$$


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?
$$\frac{6.00L}{3.15mol/L}=1.94mol$$ - Sulfuric Acid Solution

$$\frac{4.50L}{7.00mol/L}=0.64mol$$ - 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 ofH₂SO_4(aq) is 98.03g

$$1.30mol$$ x $$(\frac{98.03g}{1mol})=127.44g$$



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

 

[bross7]

1) You had the right idea but you flipped over what you needed to do. Try and keep the data organized together.

C₁V₁=C₂V₂

You have both the final concentration and volume and the initial concentration so you are trying to find the original volume.

2) This problem is asking if it is possible to make 850mL of the 1.85mol/L solution if you have 200mL of 7.50mol/L solution to start with. So this question is similar to problem 1 except this time they want you to compare moles. Try finding the number of moles initially and at the end and seeing if they are the same or not.

3a) One of the important things to do in questions is to always pay attention to units. The second part of your calculation is correct, but when you found the moles of Mg(OH)2 you found $$L^2/mol$$ instead of moles.

$$n = C*V = (2.25mol/L)(5.50L) = 12.375mol$$

3b) Same problem as in 3a

3c) Watch you work out limiting reagent problems. The reaction depends on your limiting reagent (H2SO4)

$$n_{H2SO4}=18.9mol$$

Using the mole ratio:

$$n_{H2O}=n_{H2SO4}*\frac{2}{1}$$

Finally using the molar mass of water (18.0g/mol) find the mass of water produced:

$$m=n*MM$$

 

[thepatientmental]

.


Overall, your work looks good. Here are some minor corrections/suggestions:

1. For the first question, the final answer should be in liters (L) since the units for concentration are in mol/L. So the correct answer would be 3.0 L.

2. For the second question, you can use the following equation to determine the volume of the original solution needed to make the desired dilution:
V1C1 = V2C2
Where:
V1 = volume of original solution
C1 = concentration of original solution
V2 = final volume of dilution
C2 = desired concentration of dilution

In this case, you can rearrange the equation to solve for V1:
V1 = (V2C2)/C1
Plugging in the values:
V1 = (850mL * 1.85mol/L) / 7.50mol/L = 210 mL
Since you only have 200mL of the original solution, it is not possible to make the desired dilution.

3. For part (b) of the third question, you correctly calculated the moles of each reagent, but it would be helpful to show the balanced equation to explain why the magnesium hydroxide is the limiting reagent. You also need to subtract the amount of magnesium hydroxide used in the reaction from the initial amount to determine the excess:
Initial moles of Mg(OH)2 = 7.00mol/L * 4.50L = 31.5 mol
Moles used in reaction = 5.50L * 2.25mol/L = 12.375 mol
Excess moles = 31.5 mol - 12.375 mol = 19.125 mol

Hope this helps! Good job on your work.