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. show me the steps!

              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

answer: they produce more slow than unicellular organisms.