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fredd [130]
3 years ago
7

50 points I need help on this whole work sheet about converting moles

Chemistry
1 answer:
kozerog [31]3 years ago
8 0

Answer:

Explanation:

11)

Answer:

9.08 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of O₂ = 22.7 mol

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

                               O₂      :        P₂O₅

                                 5      :          2

                                 22.7  :        2/5×22.7 = 9.08

12)

Answer:

7 mol

Given data:

Number of moles of P₂O₅ = ?

Number of moles of P = 14 mol

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with P.

                               P        :        P₂O₅

                               4        :          2

                                14      :        2/4×14 = 7

13)

Answer:

76.25 mol

Given data:

Number of moles of P =  61 mol

Number of moles of O₂ react = ?

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P with O₂.

                                  P         :        O₂

                                  4          :          5

                                 61          :        5/4×61 = 76.25

14)

Answer:

1.25 mol

Given data:

Number of moles of P₂O₅ = 0.5 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

                                P₂O₅         :        O₂

                                  2            :          5

                                0.5          :        5/2×0.5 = 1.25

15)

Answer:

20 mol

Given data:

Number of moles of P₂O₅ = 8 mol

Number of moles of O₂ needed = ?

Solution:

Chemical equation:

4P +  5O₂    →    2P₂O₅

Now we will compare the moles of P₂O₅ with O₂.

                                P₂O₅       :        O₂

                                  2            :          5

                                  8            :       5/2×8 = 20

16)

Answer:

12 mol

Given data:

Number of moles of silver made = ?

Number of moles of Ag₂O = 6 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles of Ag with Ag₂O .

                       Ag₂O      :       Ag

                           2         :        4

                           6          :        4/2×6 = 12

17)

Answer:

25 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 6.25 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles of Ag with O₂ .

                          O₂             :       Ag

                           1               :        4

                           6.25          :      4×6.25 = 25

18)

Answer:

9.8 mol

Given data:

Number of moles of silver made = ?

Number of moles of O₂ produced = 2.45 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles of Ag with O₂ .

                          O₂             :       Ag

                           1               :        4

                          2.45          :      4×2.45 = 9.8

19)

Answer:

4.4 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 2.2 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

                           O₂            :       Ag₂O

                           1               :         2

                          2.2            :        2×2.2  = 4.4

20)

Answer:

1.5 mol

Given data:

Number of moles of silver oxide required = ?

Number of moles of O₂ produced = 0.75 mol

Solution:

Chemical equation:

2Ag₂O   →   4Ag + O₂

Now we will compare the moles Ag₂O of with O₂ .

                           O₂            :         Ag₂O

                           1               :            2

                          0.75            :        2×0.75 = 1.5

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A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

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