Answer:
10 moles of SO₂ are produced when 5 moles of FeS₂
Explanation:
Stoichiometry: it is the theoretical proportion in which the chemical species are combined in a chemical reaction. The stoichiometric equation of a chemical reaction relates molecules or number of moles of all the reagents and products that participate in the reaction.
In other words, stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships established are molar relationships (that is, moles) between the compounds or elements that make up the chemical equation.
The stoichiometric coefficients of a chemical reaction indicate the proportion in which said substances react.
Taking into account the above, you can apply the following rule of three: by stoichiometry if 4 moles of FeS₂ produce 8 moles of SO₂, then when reacting 5 moles of FeS₂ how many moles of SO₂ will they produce?

moles of SO₂= 10
<u><em>10 moles of SO₂ are produced when 5 moles of FeS₂</em></u>
Answer:
a. 3-methylbutan-2-ol
b. 2-methylcyclohexan-1-ol
Explanation:
For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>
In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.
For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.
See figure 1
I hope it helps!
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