Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. the swedish chemist alfred nobel (1833-1896) founded the nobel prizes with a fortune he made by inventing dynamite, a mixture of nitroglycerin and inert ingredients that was safe to handle.

1. write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin ( c 3 h 5 no 3 3 ) into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide.

2. suppose 69.0 l of carbon dioxide gas are produced by this reaction, at a temperature of − 5.0 °c and pressure of exactly 1 atm . calculate the mass of nitroglycerin that must have reacted. be sure your answer has the correct number of significant digits.

1. 4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)

2. 238 g of C₃H₅(NO₃)₃ has been reacted.

Explanation:

This is the chemical reaction:

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

For the second part, let's apply the Ideal Gases Law to find out the moles of CO₂ that were produced.

P . V = n . R . T

1 atm . 69L = n . 0.082 L.atm/mol.K . 268K

(1 atm . 69L) / (0.082 L.atm/mol.K . 268K) = n  → 3.14 moles

In the equation, ratio between nitroglycerin and CO₂ is 12:4.

12 moles of CO₂ were produced by 4 moles of C₃H₅(NO₃)₃

Then, 3.14 moles of CO₂ would have been produced by (3.14  .4) / 12 = 1.04 moles of C₃H₅(NO₃)₃

Let's convert the moles to mass, to find out the mass of nitroglycerin that must have reacted (mol . molar mass)

1.04 mol . 227.08 g/mol = 238 g

answer; natural resources;

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o