If 4.0 mol aluminum and 7.0 mol hydrogen bromide react according to the following equation, how many moles of hydrogen are formed and what is the limiting reactant?

Moles of hydrogen formed = 3.5 moles

Explanation:

Given that:-

Moles of aluminium= 4.0 mol

Moles of hydrogen bromide = 7.0 mol

According to the reaction:-

2 moles of aluminum react with 6 moles of hydrogen bromide

1 mole of aluminum react with 6/2 moles of hydrogen bromide

4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide

Moles of hydrogen bromide = 12 moles

Available moles of hydrogen bromide = 7.0 moles

Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)

The formation of the product is governed by the limiting reagent. So,

6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen

1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen

7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen

Moles of hydrogen formed = 3.5 moles

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.  

the answer would be b i believe