1.Reaccionan 9.7 Kg de un mineral de níquel al 70% con 8L de una solución de ácido fosfórico al 60% y con una densidad de 1.36g/ml.

The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.

Explanation:

English Translation

9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.

Solution

The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.

The balanced chemical reaction between Nickel and Phosphoric acid is given as

3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂

We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.

Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg

Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg

Converting both of these to number of moles

Number of moles = (mass)/(Molar mass)

For nickel,

Mass = 6.79 kg = 6790 g

Molar mass = 58.6934 g/mol

Number of moles at the start = (6790/58.6934) = 115.7 moles

For Phosphoric acid

Mass = 6528 g

Molar mass = 97.994 g/mol

Number of moles = (6528/97.994) = 66.6 moles

3 moles of Ni reacts with 2 moles of H₃PO₄

From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.

From the stoichiometric balance of the reaction

2 moles of H₃PO₄ gives 3 moles of H₂

66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.

Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g

2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂

66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.

Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg

Hope this Helps!!!

5.6⋅10−^19  j

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