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Lorico [155]
3 years ago
12

A chemist adds of a aluminum chloride solution to a flask. Calculate the mass in grams of aluminum chloride the chemist has adde

d to the flask. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
DerKrebs [107]3 years ago
4 0

The given question is incomplete. The complete question is:

A chemist adds 95.0 ml of a 1.5M aluminum chloride solution to a flask. Calculate the mass in grams of aluminum chloride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer:  19.0 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of AlCl_3 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{133.34g/mol}

Now put all the given values in the formula of molarity, we get

1.5=\frac{x\times 1000}{95.0\times 133.34}

x=19.0g

Therefore, the mass in grams of aluminum chloride the chemist has added to the flask is 19.0

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13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
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Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
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3 years ago
A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut
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Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

  • It is a stichiometry problem.
  • We should write the balance equation of the mentioned chemical reaction:

<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>

  • It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
  • Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

<em>n = mass / molar mass</em>

  • The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
  • The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³  mol.
  • <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>

∴ 3.85 x 10⁻³  mol of Al foil reacts completely with 5.578 x 10⁻³  mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

  • From the stichiometry 3.0 moles of  CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
  • So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
  • Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³  mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

  • <u><em>So, the answer is:</em></u>

<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>

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The Nazca plate is converging with the South American plate.

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