If the CaCO3 weighed 612 g and the CaO weighed 343 g, how many grams of CO2 were formed in the reaction?
A sample of CaCO3 was heated, causing it to form CaO and CO2 gas. Solid CaO remained behind, while the CO2 escaped to the atmosphere. If the CaCO3 weighed 612 g and the CaO weighed 343 g, how many grams of CO2 were formed in the reaction?
`CaCO_(3(s)) rarr CO_(2(g)) + CaO_((s))`
`CaO:CO_2 = 1:1`
CaO = 56 g/mol
`CO_2` = 44 g/mol
`CaCO_3` = 100 g/mol
CaO moles formed when heating = (CaO weight)/(CaO molar mass)
= 6.125 mol
So number of `CO_2` moles formed = 6.125 mol
If we consider the `CaCO_3` ;
Amount of `CaCO_3` heated = 612/100
= 6.12 mol
But mole ratio of;
`CaCO_3:CO_2:CaO = 1:1:1`
So we cannot have 6.125 moles of CaO or `CO_2` . The maximum we can get is 6.12 moles. So When heating 6.12moles of `CO_2` should form.
Mass of `CO_2` = 6.12*44 g = 269.28 g
So in the reaction 269.28 g of `CO_2` will be formed.
First you need the full balanced equation
CaCO3 → CaO + CO2
You don't need to add coefficients. It is already balanced. This means one mole of CaCO3 makes one mole of CaO and one mole of CO2.
But you don't have one mole of CaCO3. You have 612g. How many moles is that? You can calculate by using the fomula
moles = actual mass/molar mass
And the molar mass of CaCO3 is 100.09g/mol.
Try this first step and see if you get it.