Question

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

Answer 1

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