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