Nitroglycerin, an explosive, decomposes according to the following equation 4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) Calculate the total volume of gases produced when collected at 1.45 atm, and 18.0°C from 2.70 × 102 g of nitroglycerin.

6.65dm³

Explanation:

Equation of reaction,

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

From the equation of reaction, 4 moles of Nitroglycerin gave 29 moles of various gases.

Molar mass of nitroglycerin C₃H₅(NO₃)₃ = 908g

Since all the product of the reaction are in gaseous phase, let's assume that law of conservation of matter is held hence there's no loss in mass.

908g of C₃H₅(NO₃)₃ = 908g of products

2.70×10²g of C₃H₅(NO₃)₃ = 2.70×10²g of products

Number of moles = mass / molar mass

Molar mass of C₃H₅(NO₃)₃ = 908g/mol

Number of moles = 2.70×10² / 908

Number of moles = 0.297 moles

But 1 mole = 22.4dm³

0.297mole = x dm³

x = (0.297 × 22.4) / 1

x = 6.65dm³

The volume of gas that'll be produced when 2.70×10²g of C₃H₅(NO₃)₃ would be 6.65dm³

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