Reaccionan 49 gr de acido nitrico , con 36 gr de hidroxido de calcio , en un proceso donde se obtienen 97 gr de nitrato de calcio mas agua . Calcular el rendimiento de la reacción -Balancear y nombrar todas las ecuaciones -Realizar cada uno de los ejercicios de rendimiento

Balanced reaction: 2HNOc + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

Limiting reactant: HNO₃

Produced calcium nitrate + Produced water = 63.8 g+ 14 g = 77.8 g

This will be the theoretical yield, so you could not produce 97 g. Something's wrong.

Explanation:

The reactants are HNO₃ and Ca(OH)₂

The products are: Ca(NO₃) and H₂O

This is a neutralization reaction: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

We determine the limting reactant (mass / molar mass, to convert to moles):

49 g / 63 g/mol = 0.778 moles of nitric acid

36 g / 74.08 g/mol = 0.486 moles of hydroxide

1 mol of hydroxide reacts with 2 moles of acid

Then, 0.486 will react with (0.486 . 2) /1 = 0.972 moles of HNO₃

We do not have enough acid, because we need 0.972 moles and we have 0.778 moles, therefore the HNO₃ is the limtiing reactant:

Ratio between the nitric acid and nitrate is 2:1, so let's propose a rule of three:

2 moles of acid can produce 1 mol of nitrate

Then, 0.778 moles of acid will produce (0.778 . 1) /2 = 0.389 moles of nitrate

We convert the moles to mass: 0.389mol . 164.08 g / 1mol = 63.8 g

Ratio between the nitric acid and water is 2:2. For 0.778 moles of acid, I can produce the same amount of water. We convert the moles to mass:

0.778 mol . 18 g/mol = 14 g

Produced calcium nitrate + Produced water = 63.8 g+ 14 g = 77.8 g

This will be the theoretical yield, so you could not produce 97 g. Something's wrong.

Percent yield = (produced yield  / theoretical yield) . 100

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