chemistry 4. Balance the following equation for the combustion of benzene: 5. Al

Question

chemistry 4. Balance the following equation for the combustion of benzene: 5. Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen. A mixture of 82.49 g of aluminum (M = 26.98 g/mol) and 117.65 g of oxygen (M = 32.00 g/mol) is allowed to react. What mass of aluminum oxide (M = 101.96 g/mol) can be formed? 6. What is the percent yield for the reaction If 164.7 g of PCl5 (M = 208.2 g/mol) are formed when 61.3 g of Cl2 (M = 70.91 g/mol) react with excess PCl3?

Explanation / Answer

Answer –

4) We are given combustion reaction

Reaction –

C6H6(l) + O2(g) ----> H2O(g) + CO2(g)

First we need count each element from both side –

   C6H6(l) + O2(g) ----> H2O(g) + CO2(g)

     C = 6                          C = 1

    H = 6                          H = 2

     O = 2                          O = 3

Now we need to balance the c first- In the reactant there are 6 C and in the product there are 1 C, so we need to 6 C in product side also, so we need to put 6 CO2

      C6H6(l) + O2(g) ----> H2O(g) + 6 CO2(g)

     C = 6                          C = 6

    H = 6                          H = 2

     O = 2                          O = 13

Now we need to balance the H- In the reactant there are 6 H and in the product there are 2 H, so we need to 6 H in product side also, so we need to put 3 H2O

     C6H6(l) + O2(g) ----> 3 H2O(g) + 6 CO2(g)

     C = 6                          C = 6

    H = 6                          H = 6

     O = 2                          O = 15

Now we need to balance the O – In the reactant there are 2 O and in the product there are 15 O, so we need to 15 in reactant side also, so we need to put 15/2 O2

     C6H6(l) + 15/2 O2(g) ----> 3 H2O(g) + 6 CO2(g)

     C = 6                                     C = 6

    H = 6                                     H = 6

     O = 15                                    O = 15

Now we need whole number, so multiply by 2 to whole

     2C6H6(l) + 15 O2(g) ----> 6 H2O(g) + 12 CO2(g)

     C = 6                                     C = 6

    H = 6                                     H = 6

     O = 15                                    O = 15

So answer is – option c

Q 5 ) We are given, mass of Al = 82.79 g , mass of O2 = 117.65 g

First we need to calculate moles of each reactant

Moles of Al = 82.79 g / 26.98 g.mol-1 = 3.068 mol

Moles of O2 = 117.65 g / 32 g.mol-1 = 3.68 mol

Now we need to calculating the limiting reactant –

From the balanced equation

4 moles of Al = 2 moles of Al2O3

So, 3.068 moles of Al = ?

= 3.068 moles of Al * 2 moles of Al2O3 / 4 moles of Al

= 1.53 moles of Al2O3

From the balanced equation

3 moles of O2 = 2 moles of Al2O3

So, 3.68 moles of Al = ?

= 3.68 moles of Al * 2 moles of Al2O3/ 3 moles of O2

= 2.45 moles of Al2O3

So moles of Al2O3 is lowest from the moles of Al, so Al is limiting reactant.

Moles of Al2O3 = 1.53 moles

Mass of Al2O3 = 1.53 moles * 101.96 g/mol

                         = 156 g

So, answer is b. 155.8 g

Q 6 ) we are given, mass of Cl2 = 61.3 g

Mass of PCl5 = 164.7 g

First we need to calculate moles of Cl2 = 61.3 g / 70.906 g.mol-1

                                                               = 0.864 moles

Now we need to calculate moles of PCl5

From the balanced equation –

1 moles of Cl2 = 1 moles of PCl5

So, 0.864 moles of Cl2 = ?

= 0.864 moles PCl5

So, theoretical yield of PCl5 = 0.864 mole * 208.2 g/mol

                                              = 179.99 g of PCl5

Percent yield = 167.7 g / 179.99 g * 100 %

                      = 93.2 %

So answer is near to e) 91.5 %

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.