How many moles of ammonia could be made given a #10*kg# mass of dihydrogen, and a #50*kg# mass of dinitrogen?

Answer 1

We assume (fancifully) quantitative reaction....

.....and we interrogate the molar quantities of each reagent in the stoichiometric equation....

#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#
#"Moles of dinitrogen"=(50*kgxx10^3*g*kg^-1)/(28.02*g*mol^-1)=1784.4*mol#
#"Moles of dihydrogen"=(10*kgxx10^3*g*kg^-1)/(2.015*g*mol^-1)=4962.8*mol#....

...clearly there is a stoichiometric EXCESS of dinitrogen with respect to the stoichiometric equation, and dihydrogen is the limiting reagent.

And so at most we could make....#2/3*"equiv"xx4962.8*mol=308.5*mol# with respect to ammonia.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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