how Glycerol C3H4O3, was combusted to carbon dioxide and water: C3H3O3 (I) + O2(

Question

how Glycerol C3H4O3, was combusted to carbon dioxide and water: C3H3O3 (I) + O2(g) rightarrow CO2(g) + H2O (I) In another reaction, glycerol underwent nitration to nitroglycerin, C2H3N3O3, Nitroglycerin is a powerful explosive Its decomposition may be represented by. This reaction generates a large amount o1 heat and many gaseous products, it is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. Balance Equation I by providing the correct whole-number coefficient in each blank. How many nitrogen atoms would be in 1.0 g of nitroglycerin? According to Equation II, how many grams of nitroglycerin will be produced if 1.05 g of glycerol is reacted with an excess of HNO3? Assume the reaction goes to completion.

Explanation / Answer

Answer -

3)We are given the combustion reaction of glycerol and we need to balance the equation

C3H8O3(l) + O2(g) ---> CO2(g) + H2O(l)

In the first step there is need for balanced C – In the reactant side there are 3 C and in the product there are 1 C, so we need to 3 C in product also

C3H8O3(l) + O2(g) ---> 3 CO2(g) + H2O(l)

C = 3                                 C = 3

H = 8                                H = 2

O = 5                                 O = 7

Now we need to balance the H

C3H8O3(l) + O2(g) ---> 3 CO2(g) + 4 H2O(l)

C = 3                                 C = 3

H = 8                                H = 8

O = 5                                 O = 10

Now we need to balance the O –

C3H8O3(l) + 7/2 O2(g) ---> 3 CO2(g) + 4 H2O(l)

C = 3                                 C = 3

H = 8                                H = 8

O = 10                                 O = 10

We need the mole coefficient in the whole number

2 C3H8O3(l) + 7 O2(g) ---> 6 CO2(g) + 8 H2O(l)

C = 6                                       C = 6

H = 16                                    H = 16

O = 20                                     O = 20

Q 4) we are given , mass of nitroglycerin = 1.0 g

We need to calculate moles of nitroglycerin

Moles of nitroglycerin = 1.0 g / 227.11 g.mol-1

                                     = 0.004403 moles

We know,

1 moles of nitroglycerin C3H5N3O9 = 3 moles of N

So, 0.004403 moles of C3H5N3O9 = ?

= 0.01321 moles

We also know,

1 mole = 6.023*1023 atoms

So, 0.01321 moles =

= 7.96*1021 atoms of N

Q 5 ) Mass of glycerol = 1.05 g ,

Reaction-

C3H8O3(l) + 3 HNO3 -------> C3H5N3O9 (s) + 3 H2O(l)

Moles of C3H8O3(l) = 1.05 g / 92.11 g.mol-1

                                 = 0.0114 moles

From the above balanced equation –

1 moles of C3H8O3(l) = 1 moles of C3H5N3O9 (s)

So, 0.0114 moles of C3H8O3 = ?

= 0.0114 moles of C3H5N3O9 (s)

Mass of C3H5N3O9 (s) = 0.0114 moles * 227.11 g/mol

                                     = 2.59 g of C3H5N3O9 (s)

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