Question

N2H4 + N2O4 --> N2 + H2O

Answer 1

Answer:

1. The limiting reagent is N2O4
2. 14,09g

Explanation:

• First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄$6N_{2} + 4H_{2}O$

• Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

               $molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

                                           $molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

              $molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

                                           $molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴                          $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

                                      $MassOfH_{2}O = 14,09g$