Nitro glycerine is used as a explosive. The equation for the explosive reaction is underset((l))(4C_(3)H_(5)((NO_(3)))_(3))rarr underset((g))(12CO_(2))+underset((l))(10H_(2)O)+underset((g))(6N_(2))+underset((g))(O_(2)) ("Atomic mass of C = 12, H = 1, N = 14, O = 16") (i) How many moles does the equation show for (a) Nitroglycerine (b) gas molecules produced? (ii) How many moles of gas molecules are obtained from 1 mole of nitroglycerine? (iii) What is the mass of 1 mole of nitroglycerine?

Nitro glycerine is used as a explosive. The equation for the explosive

1.	Nitro glycerine is used as a explosive. The equation for the explosive reaction is underset((l))(4C_(3)H_(5)((NO_(3)))_(3))rarr underset((g))(12CO_(2))+underset((l))(10H_(2)O)+underset((g))(6N_(2))+underset((g))(O_(2)) ("Atomic mass of C = 12, H = 1, N = 14, O = 16") (i) How many moles does the equation show for (a) Nitroglycerine (b) gas molecules produced? (ii) How many moles of gas molecules are obtained from 1 mole of nitroglycerine? (iii) What is the mass of 1 mole of nitroglycerine?
Answer» Solution :(i) 4 moles of Nitroglycerine (II) 4 moles of Nitroglycerine produce 19 moles of gas molecules `THEREFORE"1 mole of Nitroglycerine produces "19//4="4.75 moles"` (iii) Mass of 1 mole of Nitroglycerine `C_(3)H_(5)(No_(3))_(3)` `{:("Atomic mass of C ",=, 12),("Atomic mass of 3(C )"=3xx12,=,36),("Atomic mass of 5(H)"=5xx1,=,5),("Atomic mass of 3(N)"=3xx14,=,42),("Atomic mass of 9(O)"=9xx16,=,144),(,,""ulbar("227g")):}` Mass of 1 mole of Nitroglycerine = 227g