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balandron [24]
3 years ago
11

Anyone? Please help

Chemistry
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

The limiting reacting is O2

Explanation:

Step 1: data given

Number of moles O2 = 21 moles

Number of moles C6H6O = 4.0 moles

Step 2: The balanced equation

C6H6O + 7O2 → 6CO2 + 3H2O

Step 3: Calculate the limiting reactant

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed (21 moles).

C6H6O is in excess.

For 7 moles O2 we need 1 mol C6H6O

For 21 moles O2 we'll need 21/7 = 3 moles C6H6O

There will remain 4.0 - 3.0 = 1 mol C6H6O

Step 4: calculate products

For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O

For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2

For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O

The limiting reacting is O2

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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

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