Question

50kg of N2 and 10.0kg of H2 are mixed to produce NH3 calculate the NH3 formed and identify the limiting reagent

Answer 1

Abbreviations of 10 to the power :-

deca(D) = 10¹ , kilo (K) = 10^3 , mega or million (M) = 10^6. , giga(G) = 10^9. , tera(t) = 10^12. ,

peta(P) = 10^15 , exa(e) = 10^18. , zotta(Z) = 10^21. , yotta(Y) = 10^24

deci (d) = 10^-1. , centi(c) = 10^-2. , Milli (m) = 10^-3. , micro(µ) = 10^-6 , zocta (z) =

10^-21 , nano(n) = 10^-9 .,. fecto(f) = 10^-15 , yocta(y) = 10^-24.

some other abbreviations :-
1 angstrom = 10^-10 meter

1 Fermi = 10^-15 meter

1 light year = 9.1 × 10^15 meter

1 astronomical unit = 1.5 × 10^11 meter

1 parsec = 3.26 ly

[ just for information]
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N2 + 3H2 --------> 2NH3

(14×2=28g). + (1×2×3=6g). ---------> (14+3)×2 = 34g

50 ×10^3g + 10× 10^3g ---------> ?

to identify limiting reagents .
in balanced chemical reaction 6g of H2 requires 28 g of N2 (i.e 6×4 = 28 g)
means if 10kg ofH2 requires 40 kg of N2 (i.e 10×10^23 ×4 = 40×10^3g) so here N2 is excess reagents hence H2 is limiting reagent.
so we will only consider limiting reagent

6g of H2 gives -------> 34 g of NH3
10 kg of H2 gives -------> " X " grams of NH3.

6g : 10×10^3g :: 34g : X grams

6 × X = 10 × 10^3 × 34
X = 340 × 10^3/6
X = 56.67 × 10^3

so 56.67 × 10^3 g or 56.67 kg produced
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