Recall that the law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. This means that when a chemical reaction occurs and something new is created, the mass of the final product(s) must be equal to the mass of the original reactant(s). In other words:
Mass of reactants \( = \) Mass of products
In this example, you can see that the total mass of the reactants (CaCl2 and Na2SO4) before the reaction occurs was 184.34 grams. When they are mixed, a chemical reaction occurs and forms the products (CaSO4 and NaCl), which have a total mass of 184.34 grams. Notice that this is the same as the mass of the reactants.
Let's Practice
Practice applying the law of conservation of mass to chemical reactions by completing this activity. Answer the question on each tab, then check your answer.
What mass of rust (Fe2O3) will be produced when 25.5 grams of iron reacts with 11.0 grams of oxygen?
\( 4Fe + {3O}_{2} \rightarrow {2Fe}_{2}O_{3} \)
\( 36.5\ g \) of \( \text{Fe}_{2}O_{3} \)
If you need help arriving at this answer, click the Solution button.
Step 1: Write an equation showing conservation of mass of reactants and products. |
Mass of reactants \( = \) Mass of products Mass of \( \text{Fe} \ + \) Mass of \( O_{2} = \) Mass of \( \text{Fe}_{2}O_{3} \) |
Step 2: Substitute known values into the equation and solve for the unknown value. |
Mass of \( \text{Fe} \ + \) Mass of \( O_{2} = \) Mass of \( \text{Fe}_{2}O_{3} \) \( 25.5\ g + 11.0\ g = \) Mass of \( \text{Fe}_{2}O_{3} \) \( \require{color}\colorbox{yellow}{$ 36.5\ g $} = \) Mass of \( \text{Fe}_{2}O_{3} \) |
How many grams of table salt will be formed when 6.4 g of sodium metal reacts with 9.9 g of chlorine gas?
\( 2Na + \text{Cl}_{2} \rightarrow 2NaCl \)
\( 16.3\ g \) of \( NaCl \)
If you need help arriving at this answer, click the Solution button.
Step 1: Write an equation showing conservation of mass of reactants and products. |
Mass of reactants \( = \) Mass of products Mass of \( \text{Na} \ + \) Mass of \( \text{Cl}_{2} = \) Mass of \( \text{NaCl} \) |
Step 2: Substitute known values into the equation and solve for the unknown value. |
Mass of \( \text{Na} \ + \) Mass of \( \text{Cl}_{2} = \) Mass of \( \text{NaCl} \) \( 6.4\ g + 9.9\ g = \) Mass of \( \text{NaCl} \) \( \require{color}\colorbox{yellow}{$ 16.3\ g $} = \) Mass of \( \text{NaCl} \) |
A plant takes in 21.3 grams of carbon dioxide and 8.7 grams of water. If it produces 14.5 grams of glucose, how much oxygen gas is also produced?
\( 6\text{CO}_{2} + {6H}_{2}O \rightarrow C_{6}H_{12}O_{6} + 6O_{2} \)
\( 15.5\ g \) of \( O_{2} \)
If you need help arriving at this answer, click the Solution button.
Step 1: Write an equation showing conservation of mass of reactants and products. |
Mass of reactants \( = \) Mass of products Mass of \( \text{CO}_{2} \ + \) Mass of \( H_{2}\text{O} = \) Mass of \( C_{6}H_{12}O_{6} \ + \) Mass of \( 6O_{2} \) |
Step 2: Substitute known values into the equation and solve for the unknown value. |
Mass of \( \text{CO}_{2} \ + \) Mass of \( H_{2}\text{O} = \) Mass of \( C_{6}H_{12}O_{6} \ + \) Mass of \( 6O_{2} \) \( 21.3\ g + 8.7\ g = 14.5\ g \ + \) Mass of \( 6O_{2} \) \( 30.0\ g = 14.5\ g \ + \) Mass of \( 6O_{2} \) \( 30.0\ g - 14.5\ g = \) Mass of \( 6O_{2} \) \( \require{color}\colorbox{yellow}{$ 15.5\ g $} = \) Mass of \( 6O_{2} \) |
