Question

N the following reaction, how many grams of nitroglycerin C3H5(NO3)3 will decompose to give 120 grams of water?

Answer 1

Answer : The mass of nitroglycerin decompose is 604.1 grams.

Solution : Given,

Mass of $H_2O$ = 120 g

Molar mass of $H_2O$ = 18.0158 g/mole

Molar mass of $C_3H_5(NO_3)_3$ = 227.0995 g/mole

First we have to calculate the moles of $H_2O$.

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

Now we have to calculate the moles of $C_3H_5(NO_3)_3$

The balanced chemical reaction is,

$4C_3H_5(NO_3)_3(l)\rightarrow 12CO_2(g)+6N_2(g)+10H_2O(g)+O_2(g)$

From the balanced reaction we conclude that

As, 10 moles of $H_2O$ produced from 4 moles of $C_3H_5(NO_3)_3$

So, 6.66 moles of $H_2O$ produced from $\frac{6.66}{10}\times 4=2.66$ moles of $C_3H_5(NO_3)_3$

Now we have to calculate the mass of $C_3H_5(NO_3)_3$

$\text{ Mass of }C_3H_5(NO_3)_3=\text{ Moles of }C_3H_5(NO_3)_3\times \text{ Molar mass of }C_3H_5(NO_3)_3$

$\text{ Mass of }C_3H_5(NO_3)_3=(2.66moles)\times (227.0995g/mole)=604.1g$

Therefore, the mass of nitroglycerin decompose is 604.1 grams.

Answer 2

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

4 x ( 227.0995 ) g  ---------------> 10 x ( 18.0158) g
mass of nitroglycerin-------------> 120 g

m ( nitroglycerin ) = 120 x 4 x 227.0995 / 10 x 18.0158

m ( nitroglycerin ) = 109007.76 / 180.158

m ( nitroglycerin ) = 605.06 g

hope this helps!