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The reaction of 7.8 g benzene, C6H6, with excess HNO3 resulted in
0.90 g of H20. What is the percentage yield?
C6H6 + HNO3 - C6H5NO2 + H2O
Molar Mass: C6H6 = 78 g/mol; HNO3 = 63 g/mol; C6H5NO2 =123
g/mol; H2O =18 g/mol

Respuesta :

nataliabrito19
Molar más: C6H6 =78 g/mol ;HNO3= 63g/ mol; C6H5NO2= 123
pstnonsonjoku

50% of water was produced

The equation of the reaction is shown below;

[tex]C6H6 + HNO3 ---> C6H5NO2 + H2O[/tex]

Since HNO3  is in excess then C6H6 must be the limiting reactant.

Amount of C6H6 reacted = 7.8 g/78 g/mol = 0.1 mole

From the balanced reaction equation;

1 mole of C6H6 yields 1 mole of water

0.1 moles of C6H6 yields 0.1 * 1/1 = 0.1 moles of water

Mass of water produced = 0.1 moles of water * 18 g/mol = 1.8 g of water

% yield = actual yield/theoretical yield * 100/1

actual yield of water = 0.90 g

% yield = 0.90 g/1.8 g  * 100 = 50%

For a definition of limiting reactant see:

https://brainly.com/question/18803167

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