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.

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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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

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.