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allsm [11]
3 years ago
11

A 1.26-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of the chloride is recovered as 3

.26 g of silver chloride. Identify the metal.
Chemistry
1 answer:
erastova [34]3 years ago
5 0

Answer:

The metal is calcium (to form CaCl2)

Explanation:

Step 1: Data given

Mass of sample of an alkaline earth metal chloride = 1.26 grams

Silvernitrate (AgNO3) = in excess

All of the chloride is recovered as 3.26 g of silver chloride.

Step 2: The equation

XCl2 + 2AgNO3 → X(NO3)2+ 2 AgCl

Step 3: Calculate moles AgCl

Moles AgCl = mass AgCl / molar mass AgCl

Moles AgCl = 3.26 grams / 143.32 g/mol

Moles AgCl = 0.0227 moles

Step 4: Calculate moles XCl2

For 1 mol XCl2 we'll have 2 moles AgCl

For 2*0.0227 moles we need 0.0227/2 = 0.01135 moles

Step 5: Calculate molar mass XCl2

Molar mass XCl2 = mass / moles XCl2

Molar mass XCl2 = 1.26 grams / 0.01135 moles

Molar mass XCl2 = 111 g/mol

Step 6: Calculate of metal X

Molar mass X = molar mass XCl2 - molar mass Cl2

Molar mass X = 111 - 70.9 = 40.1 g/mol

The metal is calcium (to form CaCl2)

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Which of the following statements is true?
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Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the form
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This is the balanced equation:

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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

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