Question

At high temperatures calcium carbonate decomposes according to the equation: CaCO3(s) → CaO(s) + CO2(g) What mass of calcium oxide is formed when 10.01 g of calcium carbonate decomposes completely? At high temperatures calcium carbonate decomposes according to the equation: CaCO3(s) → CaO(s) + CO2(g) What mass of calcium oxide is formed when 10.01 g of calcium carbonate decomposes completely? @Chemistry

Answer 1

Ok so firstly you need to determine the number of moles of calcium carbonate that reacts

If you look at the periodic table or using your memory you can determine the molar mass of calcium carbonate.

So here's what you want:
n = m/M.
n = mols.
m = mass.
M = molar mass.

You already have m as 10.01g.
Now find M.
So it's CaCO3.
Looking the at periodic table:
Ca is 40.1 g/mol, C is 12.0 g/mol and O is 16.0 g/mol (but we have 3 O so multiply that by 3).

So the molar mass of CaCO3 is 40.1 + 12.0 + 3(16.0).
=100.1 g/mol.

Finding n:
n = m/M.
n = 10.1/100.1
n = 0.100 mol of calcium carbonate (CaCO3).

Next you look at the equation.
CaCO3(s) → CaO(s) + CO2(g)
You can read it like:
For every 1 mol of CaCO3 that reacts you produce 1 mol of CaO and 1 mol of CO2.

So if you have 0.100 mol of CaCO3 that means you have 0.100 mol of CaO and CO2!

The question asked for calcium oxide (CaO).

Next, you need to find the mass of CaO needed.

So look at the equation again:
n = m/M.
To find m, rearrange the equation:
m = nM
n = 0.100.

Finding M of CaO:
M = 40.1 + 16.0 = 56.1 g/mol.

So m = 0.100 x 56.1 = 5.61 g of CaO required!

Answer 2

Whoops let me rephrase:
You have 5.61 g of CaO PRODUCED not NEEDED.