We have a solution of NaOH and H₂CO₃
First, NaOH will dissociate into Na⁺ and OH⁻ ions
The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃
The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule
If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃
Reactives
-> Products
CuO
and water are products.
I
found this reaction which has CuO and water as products: decomposition of
Cu(OH)2.
Cu(OH)2
-> CuO + H2O
Stoichiometry calculus involve the mole
proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of
CuO and 1 mole of H2O are formed.
Considering
the molar masses:
Cu(OH)2
= 83.56 g/mol
CuO
= 79.545 g/mol
H2O
= 18.015 g/mol
Then:
When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.
You
should use that numbers in the rule of three:
79.545
g CuO __________18.015 g water
3.327
g CuO__________ x =3.327*18.015 /79.545 g water
x= 0.7535 g water
