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2 years ago

If 16.4 grams of copper (II) bromide react with 22.7 grams of sodium chloride, how many grams of sodium bromide are formed?

Chemistry

1 answer:

fomenos2 years ago

4 0

The amount of sodium bromide that would be formed from the reaction will be 7.5524 grams

<h3>Stoichiometric calculation</h3>

Looking at the equation of the reaction:

$CuBr_2 + 2NaCl --- > CuCl_2 + 2NaBr$

The mole ratio of CuBr2 and NaCl is 1:2.

Mole of 16.4 grams of CuBr2 = 16.4/223.37

= 0.0734 moles

Mole of 22.7 grams of NaCl = 22.7/58.44

= 0.3884 moles

Equivalent mole of NaCl = 0.1468 moles

Thus, NaCl is in excess while CuBr2 is limiting.

Mole ratio of CuBr2 and NaBr = 1:1

Mass of 0.0734 mole NaBr = 0.0734 x 102.894

= 7.5524 grams

More on stoichiometric calculation can be found here: brainly.com/question/8062886

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2 years ago

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8 0

2 years ago

A gas system has an initial number of moles of 0.693 moles with the volume unknown. When the number of moles changes to 0.928 mo

lara [203]

Answer:

The initial volume in mL is 5959.2 mL

Explanation:

As the number of moles of a gas increases, the volume also increases. Hence, number of moles and volumes are directly proportional i.e

n ∝ V

Where n is the number of moles and V is the volume

Then, n = cV

c is the proportionality constant

∴n/V = c

Hence n₁/V₁ = n₂/V₂

Where n₁ is the initial number of moles

V₁ is the initial volume

n₂ is the final number of moles

and V₂ is the final volume.

From the question,

n₁ = 0.693 moles

V₁ = ?

n₂ = 0.928 moles

V₂ = 7.98 L

Putting the values into the equation

n₁/V₁ = n₂/V₂

0.693 / V₁ = 0.928 / 7.98

Cross multiply

∴ 0.928V₁ = 0.693 × 7.98

0.928V₁ = 5.53014

V₁ = 5.53014/0.928

V₁ = 5.9592 L

To convert to mL, multiply by 1000

∴ V₁ = 5.9592 × 1000 mL

V₁ = 5959.2 mL

Hence, the initial volume in mL is 5959.2 mL

5 0

2 years ago