Question

Suppose 23.0L of carbon dioxide gas are produced by this reaction, at a temperature of 250.0 degrees C and pressure of exactly 1 atm . Calculate the mass of calcium carbonate that must have reacted

Answer 1

So, 53.65 g of calcium carbonate must have reacted. Please check the explanation below to help you remember the steps.

Step-by-step explanation:

The balanced chemical equation for the reaction of calcium carbonate (CaCO3) to produce carbon dioxide (CO2) is:

CaCO3(s) → CaO(s) + CO2(g)

From this equation, we can see that one mole of CaCO3 produces one mole of CO2.

We can use the ideal gas law, PV = nRT, to find the number of moles of CO2 produced.

Given:

P = 1 atm

V = 23.0 L

R = 0.0821 L.atm/(K.mol) (ideal gas constant)

T = 250.0 degrees C = 250.0 + 273.15 = 523.15 K

Substituting these values into the ideal gas law gives:

n = PV/RT = (1 atm x 23.0 L) / (0.0821 L.atm/(K.mol) x 523.15 K) = 0.536 moles of CO2

Since the reaction produces one mole of CO2 for every mole of CaCO3, 0.536 moles of CaCO3 must have reacted.

The molar mass of CaCO3 is 100.09 g/mol, so the mass of CaCO3 that reacted is:

mass = moles x molar mass = 0.536 moles x 100.09 g/mol = 53.65 g

So, 53.65 g of calcium carbonate must have reacted.