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3 years ago

Please help! I don’t know how to calculate reacting masses

Chemistry

1 answer:

lord [1]3 years ago

5 0

Answer:

Mass of phosphoric acid = 6.37 g

Mass of magnesium nitride = 50.5 g

Mass of iron chloride = 22.71 g

Explanation:

1 = Given data:

Mass of phosphorus = 4 g

Mass of phosphoric oxide = ?

Solution:

4P + 5O₂ → 2P₂O₅

Number of moles of phosphorus = mass / molar mass

Number of moles of phosphorus = 4 g / 31 g/mol

Number of moles of phosphorus = 0.13 mol

Mole ration of phosphorus and phosphoric acid.

P : P₂O₅

4 : 2

0.13 : 2/4× 0.13 = 0.065 mol

Mass of phosphoric acid:

Mass of phosphoric acid = moles × molar mass

Mass of phosphoric acid = 0.065 mol × 98 g/mol

Mass of phosphoric acid = 6.37 g

2 = Given data:

Mass of magnesium = 36 g

Mass of magnesium nitride = ?

Solution:

3Mg + N₂ → Mg₃N₂

Number of moles of magnesium = mass / molar mass

Number of moles of magnesium = 36 g / 24 g/mol

Number of moles of magnesium = 1.5 mol

Mole ration of magnesium and magnesium nitride.

Mg : Mg₃N₂

3 : 1

1.5 : 1/3× 1.5 = 0.5 mol

Mass of magnesium nitride:

Mass of magnesium nitride = moles × molar mass

Mass of magnesium nitride = 0.5 mol × 101 g/mol

Mass of magnesium nitride = 50.5 g

3 = Given data:

Mass of iron = 8 g

Mass of iron chloride = ?

Solution:

2Fe + 3Cl₂ → 2FeCl₃

Number of moles of iron = mass / molar mass

Number of moles of iron = 8 g / 55.85 g/mol

Number of moles of iron = 0.14 mol

Mole ration of iron and iron chloride.

Fe : FeCl₃

2 : 2

0.14 : 0.14

Mass of iron chloride:

Mass of iron chloride = moles × molar mass

Mass of iron chloride = 0.14 mol × 162.2 g/mol

Mass of iron chloride = 22.71 g

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Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

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Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

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The equilibrium constant expression for the above reaction is as follows,

Here,

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The formula to calculate solubility

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Chemistry homework question answer, step 2, image 10

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Explanation:

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<em>T is absolute temperature = 22°C + 273.15 = 295.15K</em>

<em>P is pressure = 0.961atm</em>

<em />

Replacing:

Volume SO₂ and H₂S:

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<em>9.36x10⁶ moles H₂S</em> * 0.082atmL/molK * 295.15K / 0.961atm =

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