Question

A 187.4g sample of C3H5(NO3)3 decomposes to produce carbon dioxide, water, diatomic oxygen and diatomic nitrogen.

Answer 1

Answer:

See Explanation

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!