When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction caco3(s)→cao(s)+co2(g) what is the mass of calcium carbonate needed to produce 61.0 l of carbon dioxide at stp? express your answer with the appropriate units?

Answer is: 227.3 grams of of calcium carbonate needed.
Chemical reaction: CaCO₃ → CaO + CO₂.
V(CO₂) = 61.0 L.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 61 L ÷ 22.4 L/mol.
n(CO₂) = 2.723 mol.
From chemical reaction: n(CO₂) : n(CaCO₃) = 1 : 1.
n(CaCO₃) = 2.273 mol.
m(CaCO₃) = n(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 2.273 mol · 100 g/mol.
m(CaCO₃) = 227.3 g.

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tutorareej tutorareej · Answer 2 · 2021-10-19T10:38:36+02:00

For producing 61 L of [tex]\rm CO_2[/tex], 227.23 grams of [tex]\rm CaCO_3[/tex]. is required.

Any gas occupies 22.4 L/mol space at STP.

So, 61.0 L of gas will be;

Moles of [tex]\rm CO_2[/tex] = [tex]\rm \frac{61.0}{22.4}[/tex]

Moles of [tex]\rm CO_2[/tex] = 2.723 moles

From the reaction,

1 mole of [tex]\rm CO_2[/tex] has been produced by 1 mole of [tex]\rm CaCO_3[/tex].

2.723 moles of [tex]\rm CO_2[/tex] has been produced by 2.723 mole of [tex]\rm CaCO_3[/tex].

Mass = [tex]\rm moles\;\times\;molecular\;weight[/tex]

Mass of [tex]\rm CaCO_3[/tex]. = 2.723 [tex]\times[/tex] 100

Mass of [tex]\rm CaCO_3[/tex]. = 227.23 grams.

For producing 61 L of [tex]\rm CO_2[/tex], 227.23 grams of [tex]\rm CaCO_3[/tex]. is required.

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