Law of Conservation of Mass Problems 2014WS | PDF | Magnesium ... - Free Printable

Here are the step-by-step solutions for the problems on the worksheet. The main rule to remember for all of these is the Law of Conservation of Mass: Matter cannot be created or destroyed. This means the total mass of the reactants (stuff you start with) must equal the total mass of the products (stuff you end up with).

Problem 1

Equation: S + Fe $\rightarrow$ FeS
* Given: Total Reactants = 56 g, Total Products = 88 g.
* Logic: Mass of Reactants = Mass of Products.
* (Mass of S) + (Mass of Fe) = 88 g
* We know the total mass of reactants is listed as "56g reactants" in the box, but looking closely at the diagram, it says `[Box] + 56g reactants ==> 88g products`. Wait, let's look closer. It usually lists individual masses. Let's re-read carefully: `[Box] + 56g reactants` is likely a typo in my reading or the sheet. Let's look at the standard format. Usually, it is `Mass A + Mass B = Total Mass`.
* Let's look at the text inside the box: `[ ] + 56g reactants ==> 88g products`. This phrasing is tricky. Let's assume the "56g" refers to one of the elements (likely Iron, Fe, which has an atomic mass near 56) and the box is the other element (Sulfur).
* So: Mass of Sulfur + 56 g = 88 g.
* Mass of Sulfur = 88 - 56.
* Mass of Sulfur = 32 g.

Problem 2

Equation: Ca + ZnCO$_3$ $\rightarrow$ CaCO$_3$ + Zn
* Given:
* Calcium (Ca) = 50 g
* Zinc Carbonate (ZnCO$_3$) = 150 g
* Calcium Carbonate (CaCO$_3$) = 125 g
* Zinc (Zn) = ?
* Logic: Total Mass In = Total Mass Out.
* (50 g + 150 g) = (125 g + Mass of Zn)
* 200 g = 125 g + Mass of Zn
* Mass of Zn = 200 - 125
* Mass of Zn = 75 g.

Problem 3

Experiment: Mixing solutions in a closed system (implied by conservation laws usually taught this way, though the image shows an open beaker being poured, the scale readings are what matter).
* Given:
* Initial Total Mass (Scale reading before mixing) = 300.23 g.
* Final Total Mass (Scale reading after mixing) = 300.23 g.
* One reactant mass is given as 115.9 g (CaCl$_2$ solution).
* The question asks for the value in the black box. The arrow points from the box to the Na$_2$SO$_4$ solution beaker being poured.
* Logic: The total mass of the system stays the same. The scale reads 300.23 g. This total includes the flask on the scale AND the beaker being held/poured if the whole setup was weighed, OR more likely, the problem implies that the sum of the parts equals the total measured mass.
* Let's look at the setup. The scale reads 300.23g *before* the reaction starts (left side). The items on the scale are the flask with CaCl$_2$ (115.9g) and presumably the balance accounts for the rest? No, usually in these problems: Mass of Flask + Mass of Beaker Content = Total Mass.
* Actually, a simpler interpretation: The total mass of all chemicals involved is 300.23 g.
* Mass of CaCl$_2$ solution = 115.9 g.
* Mass of Na$_2$SO$_4$ solution (the box) = Total Mass - Mass of CaCl$_2$.
* Mass of Box = 300.23 g - 115.9 g.
* Calculation:
300.23
- 115.90
-------
184.33 g.

Problem 4

Task: Calculate the mass of the product of 6.40 g of magnesium with 1.32 g of oxygen.
* Logic: Magnesium + Oxygen $\rightarrow$ Magnesium Oxide.
* Total Mass Product = Mass of Magnesium + Mass of Oxygen.
* Total Mass = 6.40 g + 1.32 g.
* Total Mass = 7.72 g.

Problem 5

Task: Calculate the mass of zinc that reacts with 4.11 g of hydrochloric acid to form 9.1 g of zinc chloride and 3.97 g of hydrogen gas.
* Equation: Zinc + Hydrochloric Acid $\rightarrow$ Zinc Chloride + Hydrogen Gas.
* Logic: Mass of Reactants = Mass of Products.
* (Mass of Zinc + 4.11 g) = (9.1 g + 3.97 g)
* First, add the products: 9.1 + 3.97 = 13.07 g.
* Now, solve for Zinc: Mass of Zinc + 4.11 = 13.07.
* Mass of Zinc = 13.07 - 4.11.
* Calculation:
13.07
- 4.11
------
8.96 g.

Problem 6

Task: Combustion of 5.00 g of Butane. Produces 4.01 g CO$_2$ and 3.55 g H$_2$O. How much oxygen was needed?
* Equation: Butane + Oxygen $\rightarrow$ Carbon Dioxide + Water.
* Logic: Mass of Reactants = Mass of Products.
* (Mass of Butane + Mass of Oxygen) = (Mass of CO$_2$ + Mass of H$_2$O)
* (5.00 g + Mass of Oxygen) = (4.01 g + 3.55 g)
* First, add the products: 4.01 + 3.55 = 7.56 g.
* Now, solve for Oxygen: 5.00 + Mass of Oxygen = 7.56.
* Mass of Oxygen = 7.56 - 5.00.
* Mass of Oxygen = 2.56 g.

Problem 7

Task: Iron combines with 4.00 g Copper(II) nitrate to form 6.01 g Iron(I) nitrate and 0.400 g copper metal. How much iron did it take?
* Equation: Iron + Copper(II) Nitrate $\rightarrow$ Iron(I) Nitrate + Copper.
* Logic: Mass of Reactants = Mass of Products.
* (Mass of Iron + 4.00 g) = (6.01 g + 0.400 g)
* First, add the products: 6.01 + 0.400 = 6.41 g.
* Now, solve for Iron: Mass of Iron + 4.00 = 6.41.
* Mass of Iron = 6.41 - 4.00.
* Mass of Iron = 2.41 g.

Problem 8 (Sample Assessment)

Reaction: 2 H$_2$O$_2$ $\rightarrow$ 2 H$_2$O + O$_2$
Given: 72 g water and 64 g oxygen produced. What mass of H$_2$O$_2$ decomposed?
* Logic: Mass of Reactant (H$_2$O$_2$) = Mass of Products (Water + Oxygen).
* Mass H$_2$O$_2$ = 72 g + 64 g.
* Mass H$_2$O$_2$ = 136 g.
* Answer Choice: b. 136 grams.

Problem 9 (Sample Assessment)

Reaction: 2 NaCl + Ca(OH)$_2$ $\rightarrow$ CaCl$_2$ + 2 NaOH
Given: 191 g NaCl reacts with 74 g Ca(OH)$_2$. 80 g NaOH is produced. What mass of CaCl$_2$ is produced?
* Logic: Mass of Reactants = Mass of Products.
* (Mass NaCl + Mass Ca(OH)$_2$) = (Mass CaCl$_2$ + Mass NaOH)
* (191 g + 74 g) = (Mass CaCl$_2$ + 80 g)
* First, add the reactants: 191 + 74 = 265 g.
* Now, solve for CaCl$_2$: 265 = Mass CaCl$_2$ + 80.
* Mass CaCl$_2$ = 265 - 80.
* Mass CaCl$_2$ = 185 g.

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Final Answer:

1. 32 g
2. 75 g
3. 184.33 g
4. 7.72 g
5. 8.96 g
6. 2.56 g
7. 2.41 g
8. b. 136 grams
9. 185 g