Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n204 and 45.0 g n2h4. some possibly useful molar masses are as follows: n2o4 92.02 g/mol, n2h4 32.05 g/mol n204) 2 n2h4(1)3 n2(g) + 4 h2o(g)

The mass of nitrogen gas produced will be 45.64 grams.

Explanation:

To calculate the number of moles, we use the equation:

   .....(1)

Given mass of

Molar mass of

Putting values in equation 1, we get:

Given mass of

Molar mass of

Putting values in  equation 1, we get:

For the given chemical reaction:

By stoichiometry of the reaction:

1 mole of reacts with 2 moles of

So, 0.543 moles of will react with = of

As, the given amount of is more than the required amount. Thus, it is considered as an excess reagent.

Hence, is the limiting reagent.

By Stoichiometry of the reaction:

1 mole of produces 3 moles of nitrogen gas

So, 0.543 moles of will produce = of nitrogen gas.

Now, calculating the mass of nitrogen gas from equation 1, we get:

Molar mass of nitrogen gas = 28.02 g/mol

Moles of nitrogen gas = 1.629 moles

Putting values in equation 1, we get:

Hence, the mass of nitrogen gas produced will be 45.64 grams.

1.solid

2.carbon dioxide

3.helium