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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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Read 2 more answers

If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass pe

NNADVOKAT [17]

First we need to calculate the number of moles of NaOH titrated.

molar concentration = number of moles / solution volume (liter)

number of moles = molar concentration × solution volume

number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles

Then we look at the chemical reaction:

CH $_{3}$ -COOH + NaOH = CH $_{3}$ COONa + H $_{2}$ O

We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.

Now we can get the mass of acetic acid:

number of moles = mass (grams) / molecular mass (g/mol)

mass = number of moles × molecular mass

mass of acetic acid = 0,01125 × 60 = 0.675 g

We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:

concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100

concentration of acetic acid = (0.674 / 7) × 100 = 9.6 %

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Andreas93 [3]

Answer:

A.)

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The first structure accurately represents ethyl ethanoate because there is an ethyl group (2 carbons) directly attached to the oxygen within the parent chain and there are 2 carbons (the eth- in ethanoate) attached to the double bonded oxygen.

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