I work these by working three "simple" stoichiometry problems and taking the smallest value of moles H2O produced.
For example, for N2H4, it is
kmoles N2H4 = 1200/30.029= 39.96
kmoles H2O = 39.96*(8/2) = 159.85
Do the same for N2O4 and (CH3)2N2H2 and take the smaller kmoles, then
kg H2O = kmoles x 18.015 = ??
The answers you posted are correct.
Remember to round to the correct number of significant figures.
The energy used to power one of the Apollo lunar missions was supplied by the following overall reaction:
2N2H4 + (CH3)2N2H2 + 3N2O4 -> 6N2 + 2CO2 + 8H2O
For the phase of the lunar module ascended from the surface of the moon, a total of 1200kg N2H4 was available to react with 1000kg (CH3)2N2H2 and 4500kg N2O4.
a. For this portion of the flight, which of the allocated components was used up first?
b. How much water, in kg, was put into the lunar atmosphere through this reaction?
Answers:
a. N2O4
b. 2.350x10^3kg H2O
I don't know how to get these answers everytime I do it I get a really big number
1 answer
1 answer