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Vsevolod [243]
2 years ago
5

-You wish to make a 0.203 M hydrochloric acid solution from a stock solution of 6.00 M hydrochloric acid. How much concentrated

acid must you add to obtain a total volume of 175 mL of the dilute solution?
mL

-In the laboratory, a student dilutes 20.3 mL of a 8.20 M hydrochloric acid solution to a total volume of 200.0 mL. What is the concentration of the diluted solution?

Concentration =
M
-How many milliliters of 9.36 M perchloric acid solution should be used to prepare 1.50 L of 0.700 M ?

mL
Chemistry
1 answer:
lubasha [3.4K]2 years ago
8 0

For all three questions, we will use the fact that

  • molarity = (moles of solute)/(liters of solution)

1) For 175 mL of solution at 0.203 M, this means that:

  • 0.203 = (moles of solute)/0.175
  • moles of solute = 0.035523 mol

Considering the hydrochloric acid solution, if we have 0.035523 mol, then:

  • 6.00 = 0.035523/(liters of solution)
  • liters of solution = 0.035523/6.00 = 0.0059205 = <u>5.92 mL (to 3 sf)</u>

<u />

2) If there is 20.3 mL = 0.0203 L, then:

  • 8.20 = (moles of solute)/0.0203
  • moles of solute = 0.16646 mol

This means that the molarity of the diluted solution is:

  • 0.16646/(0.200) = <u>0.832 M (to 3 sf)</u>

<u />

3) If we need 1.50 L of 0.700 M solution, then:

  • 0.700 = (moles of solute)/1.50
  • moles of solute = 1.05 mol

Considering the 9.36 M acid solution, from which we need 1.05 mol of perchloric acid from,

  • 9.36 = 1.05/(liters of solution)
  • liters of solution = 1.05/9.36, which is 0.11217948717949 L, or <u>112 mL (to 3 sf)</u>
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consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of
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Answer:

Explanation:

Given data:

Mass of calcium oxide = 14.4 g

Mass of carbon dioxide = 13.8 g

Actual yield of calcium carbonate = 19.4 g

Mass of calcium carbonate produced = ?

Limiting reactant = ?

Percent yield = ?

Chemical equation:

CaO + CO₂  → CaCO₃

Number of moles of CaO:

Number of moles of CaO = Mass /molar mass

Number of moles of CaO = 14.4 g / 56.1g/mol

Number of moles of CaO = 0.26 mol

Number of moles of CO₂:

Number of moles of CO₂= Mass /molar mass

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Number of moles of CO₂ = 0.31 mol

Now we will compare the moles of CaCO₃ with CO₂ and CaO.

                  CaO           :              CaCO₃

                    1               :                 1

                 0.26           :            0.26

                  CO₂           :                CaCO₃

                  1                 :                 1

                 0.31            :               0.31

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Limiting reactant:

CaO

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Mass of CaCO₃ = moles × molar mass

Mass of  CaCO₃ = 0.26 mol × 100 g/mol

Mass of  CaCO₃ =  26 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 19.4 g/ 26 g× 100

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