Answer :
175.16 grams of[tex]CaCO3[/tex]will be produced when 98.2 grams of [tex]CaO[/tex] are reacted with an excess of [tex]CO2[/tex].
The balanced chemical equation for the reaction between[tex]CaO and CO2[/tex]is:
[tex]CaO + CO2 → CaCO3[/tex]
According to the equation, one mole of[tex]CaO[/tex] reacts with one mole of [tex]CO2[/tex]to produce one mole of [tex]CaCO3[/tex].
The molar mass of [tex]CaO[/tex]is 56.08 g/mol, and the molar mass of [tex]CO2[/tex] is 44.01 g/mol. Therefore, the number of moles of [tex]CaO[/tex] present in 98.2 g can be calculated as:
moles of [tex]CaO[/tex] = mass / molar mass = 98.2 g / 56.08 g/mol = 1.75 mol
Since the reaction is with an excess of [tex]CO2[/tex], all the [tex]CaO[/tex]will react. Therefore, the number of moles of CaCO3 produced will be the same as the number of moles of [tex]CaO[/tex] used, which is 1.75 mol.
The molar mass of [tex]CaCO3[/tex]is 100.09 g/mol. Therefore, the mass of [tex]CaCO3[/tex] produced can be calculated as:
mass of [tex]CaCO3[/tex] = moles of [tex]CaCO3[/tex] × molar mass = 1.75 mol × 100.09 g/mol = 175.16 g
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