A 187.4g sample of C3H5(NO3)3 decomposes to produce carbon dioxide, water, diatomic oxygen and diatomic nitrogen. Write the balanced chemical equation for the decomposition reaction.
How many grams of carbon dioxide are produced?
How many grams of water are produced?
How many grams of diatomic oxygen are produced?
How many grams of diatomic nitrogen are produced?
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. Verify the law of conservation of mass for the decomposition of the 187.4 g of C3H5(NO3)3 .

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Explanation:

The equation of the reaction is as follws;

4 C3H5(NO3)3 (l) → 12 CO2 (g) + 10 H2O (l) + 6 N2 (g) + O2 (g)

Number of moles of C3H5(NO3)3 = 187.4g /227 g/mol = 0.82555 moles

If 4 moles of C3H5(NO3)3 yields 12 moles of CO2

0.83 moles of C3H5(NO3)3 yields 0.82555 * 12/4 = 2.47665 moles of CO2

mass of CO2 produced = 2.47665 moles * 44 g/mol = 108.95 g of CO2

4 moles of C3H5(NO3)3 yields  10 moles of water

0.83 moles of C3H5(NO3)3 yields 0.82555 * 10/4 = 2.063875 moles of water

Mass of water = 2.063875 moles of water * 18 g/mol = 37.16 g

4 moles of C3H5(NO3)3 yields 1 mole of O2

0.83 moles of C3H5(NO3)3  yields 0.82555 * 1/4 = 0.2063875 moles of O2

Mass of O2 = 0.2063875 moles of O2 * 32 g/mol = 6.6 g

4 moles of C3H5(NO3)3 yields 6 moles of N2

0.83 moles of C3H5(NO3)3 yields 0.82555 * 6/4 = 1.238325 moles of N2

Mass of N2= 1.238325  moles of N2  *  28 g/mol = 34.67 g

To verify the law of conservation of mass, the total sum of mass of products must equal the sum of the mass of reactants. Thus;

108.95 g + 37.16 g +  6.6 g + 34.67 g = 187.4 g

The law of conservation of mass has been proven!

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