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The following data was collected when a reaction was performed experimentally in the laboratory.

reaction data
reactants products
al(no3)3 nacl nano3 alcl3
starting amount in reaction 4 moles 9 moles ? ?

required:
determine the maximum amount of nan03 that was produced during the experiment explain how you determined this amount?

Respuesta :

Eduard22sly

The maximum amount of NaNO₃ that was produced during the experiment is 12 moles

Balanced equation

Al(NO₃)₃ + 3NaCl --> 3NaNO₃ + AlCl₃

From the balanced equation above,

1 mole of Al(NO₃)₃ reacted with 3 moles of NaCl to produce 3 moles of NaNO₃

How to determine the limiting reactant

From the balanced equation above,

1 mole of Al(NO₃)₃ reacted with 3 moles of NaCl

Therefore,

4 moles of Al(NO₃)₃ will react with = 4 × 3 = 12 moles of NaCl

From the above calculation, we can see that a higher amount of NaCl (i.e 12 moles) than what was given (i.e 9 moles) are needed to react completely with 4 moles of Al(NO₃)₃.

Thus, Al(NO₃)₃ is the limiting reactant and NaCl is the excess reactant.

How to determine the maximum amount of NaNO₃ produced

To obtain the maximum amount of NaNO₃ produced, the limiting reactant (i.e Al(NO₃)₃) will be used. This is illustrated below:

Al(NO₃)₃ + 3NaCl --> 3NaNO₃ + AlCl₃

From the balanced equation above,

1 mole of Al(NO₃)₃ reacted to produce 3 moles of NaNO₃

Therefore,

4 moles of Al(NO₃)₃ will react with = 4 × 3 = 12 moles of NaNO₃

Thus, the maximum amount of NaNO₃ produced from the experiment is 12 moles

Learn more about stoichiometry:

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