How many grams of #CaO# can be formed from the decomposition of 275 g of #CaCO_3#?

Question

Avery Adams · Accepted Answer

Approx. #150*g# #"calcium oxide"# will be isolated given stoichiometric reaction.

(i) A stoichiometric equation is required. #CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)# And (ii) equivalent quantities of #"calcium carbonate"#, The formula for "moles of calcium carbonate" is (275*g)/(100.09*g*mol^-1) = 2.75*mol. And thus #2.75*molxx56.08*g*mol^-1=??*g# #"calcium oxide"#. Why is there #"1:1"# equivalence between #"calcium oxide"# and #"calcium carbonate"#.

Henry Antrim · Answer

undefined To find the number of grams of CaO formed from the decomposition of 275 g of CaCO3, you need to use stoichiometry. The balanced chemical equation for the decomposition of CaCO3 is:

CaCO3(s) -> CaO(s) + CO2(g)

From the equation, you can see that 1 mole of CaCO3 produces 1 mole of CaO. The molar mass of CaCO3 is approximately 100.09 g/mol, and the molar mass of CaO is approximately 56.08 g/mol.

Using stoichiometry:

1 mole of CaCO3 = 1 mole of CaO
100.09 g of CaCO3 = 56.08 g of CaO

Now, you can set up a proportion:

(56.08 g CaO / 100.09 g CaCO3) = (x g CaO / 275 g CaCO3)

Solving for x, you find:

x = (56.08 g CaO / 100.09 g CaCO3) * 275 g CaCO3

x ≈ 154.15 g CaO

So, approximately 154.15 grams of CaO can be formed from the decomposition of 275 grams of CaCO3.