Shays21
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In the following reaction, how many grams of nitroglycerin C3H5(NO3)3 will decompose to give 25 grams of CO2?

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


The molar mass of nitroglycerin is 227.0995 grams and that of carbon dioxide is 44.01 grams.

Verified Answer :

JoshEast

4C₃H₅(NO₃)₃\(_{(l)}\) ------> 12CO₂\(_{(g)}\) + 6N₂\(_{(g)}\) +  10H₂O\(_{(g)}\)  +  O₂\(_{(g)}\)

mol of CO₂  =  \(\frac{mass}{molar mass}\)

                    =  \(\frac{25g}{44.01 g/mol}\)

mol ratio of  CO₂ :  C₃H₅(NO₃)₃

                    12    :     4

∴  if  mole of CO₂  =  0.568 mol

then   "       "   C₃H₅(NO₃)₃  =  \(\frac{0.568 mol}{12}  *  4\)

                                           = 0.189 mol


∴ mass of nitroglycerin  =  mole    Mr

                                       =  0.189 mol    227.0995 g / mol

                                       =  43.00 g

nutanraj654

First, we have to calculate the moles of

Moles of CO2 = mass of CO2/Molar mass of CO2 = 25g/44.01g/mole =0.57 moles

Now we have to calculate the moles of nitroglycerin.

The balanced chemical reaction is,

4C3H5(NO3)3(l)~ 12CO2(G) + 6N2(g) +)2(g)

From the balanced reaction, we conclude that

As 12 moles of CO2 were obtained from 4 moles C3H5(NO3)3 of

So, 0.57 moles of CO2 were obtained from 4/12 x 0.57 = 0.19 the mole of

C3H5(NO3)3

Now we have to calculate the mass of nitroglycerin.

Mass ofC3H5 (NO3)3 = Moles of C3H5(NO3)3 x Molar masse of C3H5(NO3)3

Mass ofC3H5 (NO3)3 = (0.19mole) x (227.0995g/mole) = 43.15g

Therefore, the mass of nitroglycerin will be, 43.15 grams.

What's the decomposition of nitroglycerin?

Nitroglycerin is extremely sensitive to shock and to rapid heating; it begins to decompose at 50–60 °C (122–140 °F) and explodes at 218 °C (424 °F). The safe use of nitroglycerin as a blasting explosive became possible after the Swedish chemist Alfred B.

Learn more about  nitroglycerin at

https://brainly.com/question/4985098

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