Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
Answer:
a. Zinc is the limiting reactant.
b. 
c. 
Explanation:
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a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

Thus, since zinc has the fewest moles we infer it is the limiting reactant.
b. Here, we compute the grams of zinc bromide via both reactants:

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.
c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

Thus, the leftover of bromine is:

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Answer:
After 30 cycles, as many as a billion copies of the target sequence are produced from a single starting molecule