what volume of carbon dioxide will be collected at stp when 25.6 g of calcium carbohydrate undergoes decomposition? Balanced reaction: CaCO2 -> CaO + CO2

V = 5.74 L

Explanation:

Given data:

Volume of carbon dioxide produced = ?

Temperature and pressure = standard

Mass of calcium carbonate = 25.6 g

Solution:

Chemical equation:

CaCO₃      →        CaO + CO₂

Number of moles of calcium carbonate:

Number of moles = mass / molar mass

Molar mass of calcium carbonate = 100 g/mol

Number of moles = 25.6 g/ 100 g/mol

Number of moles = 0.256 mol

Now we will compare the mole of calcium carbonate with carbon dioxide.

                       CaCO₃           :            CO₂

                            1                :             1

                         0.256          :         0.256

Volume of  CO₂:

PV = nRT

R = general gas constant = 0.0821 atm/L / mol.K

By putting values,

1 atm× V = 0.256 mol × 0.0821 atm/L / mol.K× 273.15 K

V = 5.74 atm/L / 1 atm

V = 5.74 L

i wish i could but this one is weird give me abc and i can solve it.