For the reaction n2 + 3h2 = 2nh3, if you have 2.00 moles of n2 and 4.00 moles of h2 : a. identify the limiting reactant. b. how many moles of nh3 can be formed? c. how many moles of the excess reactant will remain?
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Explanation:

For the reaction N2 + 3H2 = 2NH3, if you have 2.00 moles of N2 and 4.00 moles of H2 : a. Identify the limiting reactant.

                                   N2     +     3H2      =      2NH3

                                 1 mol         3 moles        2 moles

Given                       2 moles     4 moles

       Rule of three for N2

                    x =  (2 x 3) / 1 = 6 mol  of H2, it means that is needed 6 moles of H2 and there are only 4, so the limiting reactant is H2                                

b. How many moles of NH3 can be formed?

           rule of three

                                3 moles of H2  2 moles of NH3              

                                4 moles of H2     x

                     x = (4x2)/3 = 2.7 moles of NH3 are formed

c. How many moles of the excess reactant will remain?

      x = (4 x 1) / 3     =  1.33

Excess reactant = 2 - 1.33 = 0.67 moles                  

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j