3. The energy used to power one of the Apollo lunar missions was supplied by the following overall reaction: 2N, H. + 2(CH3)2N, H2 + 51,04 → 9N2 + 4CO2 + 12H,0. For the phase of
the mission when the lunar module ascended from the surface of the moon, a total of
1200. kg of N He was available to react with 1000. kg of (CH3)2N_H, and 4500. kg of
N2O4
a. For this portion of the flight, which of the allocated components was used up
first?
b. How much water, in kilograms, was put into the lunar atmosphere through this
reaction?
C. How many kilograms of the of excess reactants remained upon completion of the
reaction?

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Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.