2n2h4(g)+ n2o4(g) à3 n2(g) + 4 h2o(g) when 8.0 g of n2h4(32 g mol-1) and 18.4 g of n2o4(92 g mol-1) are mixed together and react according to the equation above, what is the maximum mass of h2o that can be produced?

The mass of water that can be produced is 9.22 grams

Explanation:

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

    .....(1)

Given mass of = 8.2 g

Molar mass of = 32 g/mol

Putting values in equation 1, we get:

Given mass of = 18.4 g

Molar mass of = 92 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

2 moles of reacts with 1 mole of

So, 0.256 moles of will react with = of

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

Thus, is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of produces 4 moles of water

So, 0.256 moles of will produce = of water

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

Molar mass of water = 18 g/mol

Moles of water = 0.512 moles

Putting values in equation 1, we get:

Hence, the mass of water that can be produced is 9.22 grams

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