In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produced. given that 3.327 g of cuo was produced during the reaction, how many grams of water were released as water vapor? (number of moles = mass (g) / molar mass (g/mol)). choose the closest answer.

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

rohitdip2808 rohitdip2808 · Answer 2 · 2021-10-06T15:59:30+02:00

Answer:

The vaporized water during the reaction is 0.753g.

Explanation:

The mentioned decomposition reaction is;

[tex]\rm CuSO_2\rightarrow CuO+H_2O[/tex]

This means 1mol( 83.56 g) copper hyposulfite decomposes to form 1mol(79.545g) of Copper oxide and 1mol(18.015g) of water.

If 3.327g of CuO is producing which is 26.905 times less than its malar value.

Hence, Water released during the reaction is

[tex]\rm \frac{18.015}{23.908} = 0.753[/tex]

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