How many grams of Ca and CaO were in 4 g of the mixture if 0.448 liters of hydrogen

| April 30, 2021 | education

How many grams of Ca and CaO were in 4 g of the mixture if 0.448 liters of hydrogen were released when water was added to the mixture?

To solve this problem, we write down the given: m (mixture) = 4 g, as a result of interaction with water (H2O), a gas with a volume of 0.448 l was formed
Find: m (Ca) -? and m (CaO) -?
Decision:
When calcium oxide interacts with water, hydrogen will not be formed.
Let’s write the equation of the reaction of interaction of calcium with water.
Ca + H2O = Ca (OH) 2 + H2
Let’s arrange the coefficients.
Ca + 2H2O = Ca (OH) 2 + H2
Let’s calculate the mass of calcium that has reacted.
Over hydrogen we write 0.448 l, and under hydrogen we write a constant molar volume, which is 22.4 l / mol.
Above calcium we write x g, and under calcium its molar mass, which is 40 g / mol.
Let’s compose and solve the proportion:
x = 40 * 0.448 / 22.4 = 0.8 g
Let’s calculate the mass of calcium oxide that is included in the mixture.
m (CaO) = m (mixtures) – m (Ca)
m (CaO) = 4 – 0.8 = 3.2 g
Answer: m (Ca) = 0.8 g, m (CaO) = 3.2 g



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