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Hydrazine (N2H4) emits a large quantity of
energy when it reacts with oxygen, which has
led to its use as a fuel for rockets,
N2H4(ℓ) + O2(g) → N2(g) + 2 H2O(g).
How many moles of water are produced
when 28 g of pure hydrazine is ignited in the
presence of 73 g of pure oxygen?
Answer in units of mol.

Please show me the steps!

Respuesta :

transitionstate

Answer:

              1.75 moles of H₂O

Solution:

The Balance Chemical Equation is as follow,

                                   N₂H₄  +  O₂    →    N₂  +  2 H₂O

Step 1: Calculate the Limiting Reagent,

According to Balance equation,

             32.04 g (1 mol) N₂H₄ reacts with  =  32 g (1 mol) of O₂

So,

                   28 g of N₂H₄ will react with  =  X g of O₂

Solving for X,

                       X  =  (28 g × 32 g) ÷ 32.04 g

                       X  =  27.96 g of O₂

It means 29 g of N₂H₄ requires 47.96 g of O₂, while we are provided with 73 g of O₂ which is in excess. Therefore, N₂H₄ is the limiting reagent and will control the yield of products.

Step 2: Calculate moles of Water produced,

According to equation,

            32.04 g (1 mol) of N₂H₄ produces  =  2 moles of H₂O

So,

                        28 g of N₂H₄ will produce  =  X moles of H₂O

Solving for X,

                      X  =  (28 g × 2 mol) ÷ 32.04 g

                      X  =  1.75 moles of H₂O

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