Calculate and compare the mass of the reactants with the mass of the products for each equation.
a)2KClO3 (s) ==> 2KCl(s) + 3O2 (g)
b)4NH3(g) + 6NO(g) ==> 5N2(g) + 6H2O(g)
c)4K(s) + O2(g) ==> 2K2O(s)
Show that each balanced equation obeys the law of conservation of mass.
Good work
To calculate and compare the mass of the reactants with the mass of the products for each equation, we need to determine the molar masses of each compound involved in the reaction.
a) 2KClO3 (s) ==> 2KCl (s) + 3O2 (g)
The molar mass of KClO3 can be calculated as:
Molar mass of K = 39.10 g/mol
Molar mass of Cl = 35.45 g/mol
Molar mass of O = 16.00 g/mol
2 moles of K = 2 × 39.10 g/mol = 78.20 g
2 moles of Cl = 2 × 35.45 g/mol = 70.90 g
3 moles of O = 3 × 16.00 g/mol = 48.00 g
Therefore, the total mass of the reactants is:
78.20 g + 70.90 g + 48.00 g = 197.10 g
Now, let's calculate the mass of the products:
2 moles of KCl = 2 × (39.10 + 35.45) g/mol = 148.10 g
3 moles of O2 = 3 × (2 × 16.00) g/mol = 96.00 g
Therefore, the total mass of the products is:
148.10 g + 96.00 g = 244.10 g
b) 4NH3 (g) + 6NO (g) ==> 5N2 (g) + 6H2O (g)
The molar mass of NH3 can be calculated as:
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
4 moles of N = 4 × 14.01 g/mol = 56.04 g
12 moles of H = 12 × 1.01 g/mol = 12.12 g
Therefore, the total mass of the reactant is:
56.04 g + 12.12 g = 68.16 g
The molar mass of N2 can be calculated as:
Molar mass of N = 14.01 g/mol
5 moles of N2 = 5 × (2 × 14.01) g/mol = 140.10 g
6 moles of H2O = 6 × (2 × 1.01 + 16.00) g/mol = 216.18 g
Therefore, the total mass of the products is:
140.10 g + 216.18 g = 356.28 g
c) 4K (s) + O2 (g) ==> 2K2O (s)
The molar mass of K can be calculated as:
Molar mass of K = 39.10 g/mol
4 moles of K = 4 × 39.10 g/mol = 156.40 g
The molar mass of K2O can be calculated as:
Molar mass of K = 39.10 g/mol
Molar mass of O = 16.00 g/mol
2 moles of K2O = 2 × (2 × 39.10 + 16.00) g/mol = 166.20 g
Therefore, the total mass of the reactants is:
156.40 g + 32.00 g = 188.40 g
Similarly, the total mass of the products is also 188.40 g.
In each balanced equation, the total mass of the reactants is equal to the total mass of the products, indicating the law of conservation of mass is obeyed.
To calculate and compare the mass of the reactants with the mass of the products, we need to consider the stoichiometry of each balanced equation. The stoichiometry tells us the molar ratio between the reactants and products, allowing us to determine the mass relationship.
a) 2KClO3 (s) ==> 2KCl (s) + 3O2 (g)
To calculate the mass of the reactants and products, we need to know the molar masses of each substance:
- Molar mass of KClO3 = 39.10 g/mol (K) + 35.45 g/mol (Cl) + 3(16.00 g/mol) (O) = 122.55 g/mol
- Molar mass of KCl = 39.10 g/mol (K) + 35.45 g/mol (Cl) = 74.55 g/mol
- Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol
In this equation, the molar ratio between KClO3 and KCl is 1:1 and the molar ratio between KClO3 and O2 is 2:3. This means that for every 2 moles of KClO3, we get 2 moles of KCl and 3 moles of O2.
Let's assume we start with 100 grams of KClO3. To find the mass of KCl, we multiply the molar mass of KCl by the number of moles of KCl (which is the same as the number of moles of KClO3 since the ratio is 1:1):
Mass of KCl = 74.55 g/mol * (2 mol / 122.55 g/mol) * 100 g = 121.11 g
To find the mass of O2, we multiply the molar mass of O2 by the number of moles of O2 (which is the same as the number of moles of KClO3, multiplied by the ratio of O2 to KClO3):
Mass of O2 = 32.00 g/mol * (3 mol / 122.55 g/mol) * 100 g = 78.22 g
The total mass of the products is therefore:
Mass of products = Mass of KCl + Mass of O2 = 121.11 g + 78.22 g = 199.33 g
The total mass of the reactants (2KClO3) is 100 g, which is less than the total mass of the products (199.33 g). This equation does not obey the law of conservation of mass.
b) 4NH3 (g) + 6NO (g) ==> 5N2 (g) + 6H2O (g)
- Molar mass of NH3 = 3(1.01 g/mol) + 14.01 g/mol = 17.03 g/mol
- Molar mass of NO = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
- Molar mass of N2 = 2(14.01 g/mol) = 28.02 g/mol
- Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
The molar ratio between NH3 and N2 is 4:5, and the molar ratio between NH3 and NO is 4:6.
Let's assume we start with 100 grams of NH3. To find the mass of N2, we multiply the molar mass of N2 by the number of moles of N2:
Mass of N2 = 28.02 g/mol * (5 mol / 17.03 g/mol) * 100 g = 82.27 g
To find the mass of H2O, we multiply the molar mass of H2O by the number of moles of H2O:
Mass of H2O = 18.02 g/mol * (6 mol / 17.03 g/mol) * 100 g = 66.73 g
The total mass of the products is therefore:
Mass of products = Mass of N2 + Mass of H2O = 82.27 g + 66.73 g = 149 g
The total mass of the reactants (4NH3 + 6NO) is 100 g, which is equal to the total mass of the products (149 g). This equation obeys the law of conservation of mass.
c) 4K (s) + O2 (g) ==> 2K2O (s)
- Molar mass of K = 39.10 g/mol
- Molar mass of O2 = 32.00 g/mol
- Molar mass of K2O = 2(39.10 g/mol) + 16.00 g/mol = 94.20 g/mol
The molar ratio between K and K2O is 4:2, and the molar ratio between O2 and K2O is 1:2.
Let's assume we start with 100 grams of K. To find the mass of K2O, we multiply the molar mass of K2O by the number of moles of K2O:
Mass of K2O = 94.20 g/mol * (2 mol / 39.10 g/mol) * 100 g = 120.72 g
The total mass of the products is therefore:
Mass of products = Mass of K2O = 120.72 g
The total mass of the reactants (4K + O2) is 100 g, which is less than the total mass of the products (120.72 g). This equation does not obey the law of conservation of mass.
In summary, equation b) obeys the law of conservation of mass because the mass of the reactants is equal to the mass of the products. However, equations a) and c) do not obey the law of conservation of mass because the mass of the reactants is less than the mass of the products.
Calculate molar masses and show that they add up on both sides. For example, the first one.
KClO3 is 122.549 and 2 moles would be
2 x 122.549 = 245.098.
On the right we have
2 x 74.551 = 149.102 for 2KCl
3 x 31.9988 = 95.9964
149.102+95.9964 = 245.098
so the product decomposes but we didn't lose anything. The second one is done the same way.