Answer :
175.17 grams of CaCO₃ are produced when 98.2 grams of CaO are reacted with an excess of CO₂ according to the given equation.
To solve this problem, we need to use stoichiometry which deals with the quantitative relationships between reactants and products in chemical reactions.
The balanced chemical equation for the reaction is:
CaO + CO₂ → CaCO₃
This equation tells us that for every 1 mole of CaO and 1 mole of CO₂ that react, we get 1 mole of CaCO₃.
We are given the mass of CaO that is used in the reaction. To calculate the mass of CaCO₃ that is produced, we need to use stoichiometry and the molar mass of CaCO₃.
The molar mass of CaCO₃ is the sum of the atomic masses of one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). Using the values from the periodic table, we can calculate the molar mass of CaCO₃ as:
molar mass of CaCO₃ = 1 × atomic mass of Ca + 1 × atomic mass of C + 3 × atomic mass of O
= 1 × 40.08 g/mol + 1 × 12.01 g/mol + 3 × 16.00 g/mol
= 100.09 g/mol
To calculate the number of moles of CaO that reacted, we can use the following equation:
n = m/M
where n is the number of moles of CaO, m is the mass of CaO, and M is the molar mass of CaO.
Using the given values, we get:
n = 98.2 g / 56.08 g/mol = 1.749 mol
This is the number of moles of CaO that reacted in the reaction.
Since the reaction is 1:1, meaning that one mole of CaO reacts with one mole of CO₂ to produce one mole of CaCO₃, we know that the number of moles of CaCO₃ produced is also 1.749 mol.
Finally, to calculate the mass of CaCO₃ produced, we can use the following equation:
m = n × M
where m is the mass of CaCO₃ produced, n is the number of moles of CaCO₃ produced, and M is the molar mass of CaCO₃.
Using the given values, we get:
m = 1.749 mol × 100.09 g/mol = 175.17 g
Learn more about CaCO₃ at https://brainly.com/question/30594488.
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