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In the following reaction, how many grams of nitroglycerin C3H5(NO3)3 will decompose to give 120 grams of water?
4C3H5(NO3)3(l) 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g)

The molar mass of nitroglycerin is 227.0995 grams and that of water is 18.0158 grams.

Respuesta :

12 moles CO2 are made by 4 moles nitroglycerin
1 mole CO2 is made by 4/12 moles nitroglycerin
0.5682 moles CO2 are made by 0.5682*4/12 moles nitroglycerin = 0.18939 moles.
0.18939 moles nitroglycerin are 0.18939*227.0995 g = 43.011 g

Answer : The mass of nitroglycerin is, 604.99 grams

Solution : Given,

Mass of water = 120 g

Molar mass of water = 18.0158 g/mole

Molar mass of [tex]C_3H_5(NO_3)_3[/tex] = 227.0995 g/mole

First we have to calculate the moles of water.

[tex]\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{120g}{18.0158g/mole}=6.66moles[/tex]

Now we have to calculate the moles of [tex]C_3H_5(NO_3)_3[/tex]

The given balanced reaction is,

[tex]4C_3H_5(NO_3)_3(l)\rightarrow 12CO_2(g)+6N_2(g)+10H_2(g)+O_2(g)[/tex]

From the reaction, we conclude that

As, 10 moles of water obtained from 4 moles of [tex]C_3H_5(NO_3)_3[/tex]

So, 6.66 moles of water obtained from [tex]\frac{4}{10}\times 6.66=2.664[/tex] moles of [tex]C_3H_5(NO_3)_3[/tex]

Now we have to calculate the mass of [tex]C_3H_5(NO_3)_3[/tex]

[tex]\text{ Mass of }C_3H_5(NO_3)_3=\text{ Moles of }C_3H_5(NO_3)_3\times \text{ Molar mass of }C_3H_5(NO_3)_3[/tex]

[tex]\text{ Mass of }C_3H_5(NO_3)_3=(2.664moles)\times (227.0995g/mole)=604.99g[/tex]

Therefore, the mass of nitroglycerin is, 604.99 grams

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