Question

5- Given the reaction
4FeS2 + 11 O2 → 2Fe2O3 + 8SO2
What mass of SO2 is produced from reaction of 3.8 g of FeS2 and excess O2?
6- What mass of SO3 forms from the reaction of 4.1 g of SO2 with an excess of O2?
7-
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
Use the equation to answer the following questions:
a)10 grams of glucose (C6H12O6) react in a combustion reaction. How many grams of each product are produced?
b) Find the number of molecules of CO2 produced. c) Find the number of molecules of water produced.

Answer 1

Answer:

Explanation:

5)

Given data:

Mass of FeS₂ = 3.8 g

Mass of SO₂ = ?

Solution:

Chemical equation:

4FeS₂ + 11O₂      →      2Fe₂O₃ + 8SO₂

Number of moles of FeS₂:

Number of mole = mass / molar mass

Number of mole = 3.8 g / 119.98 g/mol

Number of moles = 0.03 mol

Now we will compare the moles of SO₂ with FeS₂.

                             FeS₂           :            SO₂

                                 4             :             8

                                0.03        :         8/4 ×0.03 = 0.06

Mass of SO₂:

Mass = number of moles × molar mass

Mass = 0.06 mol × 64.066 g/mol

Mass = 3.84 g

6)

Given data:

Mass of SO₃ form = ?

Mass of SO₂ react = 4.1 g

Solution:

Chemical equation:

2SO₂ + O₂   →    2SO₃

Number of moles of SO₂ :

Number of moles = mass/ molar mass

Number of moles = 4.1 g/ 64.066 g/mol

Number of moles = 0.064 mol

Now we will compare the moles of SO₃  with SO₂.

                        SO₂         :          SO₃

                          2            :           2

                      0.064        :        0.064

Mass of SO₃:

Mass = number of moles × molar mass

Mass = 0.064 mol × 80.066 g/mol

Mass = 5.12 g

7)

a)

Given data:

Mass of glucose = 10 g

Mass of CO₂ formed = ?

Mass of H₂O formed = ?

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂  →      6CO₂ + 6H₂O

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 10 g/ 180.156 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of sugar with each product.

                 C₆H₁₂O₆           :            CO₂

                      1                  :               6

                     0.06             :          6×0.06 = 0.36 mol

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 0.36 mol × 44 g/mol

Mass = 15.84 g

                  C₆H₁₂O₆          :            H₂O

                      1                  :               6

                     0.06             :          6×0.06 = 0.36 mol

Mass of H₂O:

Mass = number of moles × molar mass

Mass = 0.36 mol × 18 g/mol

Mass = 6.48 g

b)

Number of moles of CO₂ = 0.36 mol

Number of moles of H₂O = 0.36 mol

Number of molecules of CO₂ = ?

Number of molecules of  H₂O = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1 mole = 6.022 × 10²³ molecules of water

0.36 mol  × 6.022 × 10²³ molecules of water  / 1 mol

2.17 × 10²³ molecules of water

For CO₂:

1 mole = 6.022 × 10²³ molecules of CO₂

0.36 mol  × 6.022 × 10²³ molecules of CO₂  / 1 mol

2.17 × 10²³ molecules of CO₂

Answer 2

The number of molecules of each compound produced is 2  × 10^23 molecules.

The equation of the reaction is;

4FeS2 + 11 O2 → 2Fe2O3 + 8SO2

Number of moles of  FeS2  = 3.8 g /119.98 g/mol = 0.032 moles

If 4 moles of  FeS2   yields 8 moles of SO2

0.032 moles of FeS2 yields 0.032 moles × 8 moles/4 moles = 0.064 moles

Mass of SO2 = 0.064 moles × 64 g/mol = 4.1 g of SO2

From;

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

Number of moles of glucose = 10 g/180 g/mol = 0.056 moles

If 1 mole of glucose produces 6 moles of each product

0.056 moles of glucose produces 0.056 moles × 6 moles /1 mole  

= 0.34 moles

Mass of CO2 = 0.34 moles × 44 g/mol = 14.96 g

Mass of H2O = 0.34 moles × 18 g/mol = 6.12 g

Number of molecules of CO2 produced =  0.34 moles × 6.02 × 10^23 molecules = 2  × 10^23 molecules

Number of molecules of water = 0.34 moles × 6.02 × 10^23 molecules = 2  × 10^23 molecules

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