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Fittoniya [83]
3 years ago
12

Trisha and her lab partner were conducting a variety of experiments to produce gases: hydrogen, oxygen, and carbon dioxide. In o

ne experiment, they added a piece of magnesium ribbon to 10 milliliters of hydrochloric acid. They observed bubbles being produced and did a variety of tests to identify the escaping gas; it proved to be hydrogen The reaction is represented by the following equation: Mg + 2HCl → MgCl2 + H2(g) Trisha and her lab partner filled a beaker with 150 milliliters of a 0.10 M HCl solution. How many moles of HCl are in the beaker? Remember, molarity = moles liter . A) 0.015 mol HCl B) 0.150 mol HCl C) 6.67 mol HCl D) 15 mol HCl
Chemistry
1 answer:
GaryK [48]3 years ago
7 0

<u>Given:</u>

Volume of HCl = 150 ml

Molarity of HCl = 0.10 M

<u>To determine:</u>

The # moles of HCl

<u>Explanation:</u>

The molarity of a solution is the number of moles of a solute dissolved in a given volume

In this case:

Molarity of HCl = moles of HCl/volume of the solution

moles of HCl = Molarity * volume = 0.10 moles.L-1 * 0.150 L = 0.015 moles

Ans: A)

Moles of HCl is 0.015

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Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

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Answer:

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