Concept explainers
(a)
Interpretation:
For the given reaction, limiting reactant is to be determined.
Concept introduction:
Lithium react with bromine spontaneously to produce lithium bromide.
Lithium is a metal. Bromine is a gas and non-metal.
(a)
Answer to Problem 82A
Br2 is limiting reactant.
Explanation of Solution
First convert the reactant in moles.
Using relation
Molar mass of lithium is 6.93 g/mol
Molar mass of Bromine gas is 159.82 g/mol
According to stoichiometry of reaction, 2 moles of lithium require 1 mole of bromine gas for complete reaction. And here Li is present in excess and left behind after completion of reaction. So, Br2 is limiting reactant.
(b)
Interpretation:
Mass of lithium bromide produced to be determined
Concept introduction:
Lithium react with bromine spontaneously to produce lithium bromide.
(b)
Answer to Problem 82A
27.09 g of LiBr
Explanation of Solution
From reaction 1 mole of Br2 form 2 moles of LiBr.
0.156 mole of Br2 form 2 times the mole of LiBr.
= 0.312 mole of LiBr
Molar mass of LiBr is 86.85g/mol
Mass of LiBr formed is:
(c)
Interpretation:
Excess reactant and excess mass to be determined.
Concept introduction:
Lithium react with bromine spontaneously to produce lithium bromide.
(c)
Answer to Problem 82A
3.288 mole of Li is in excess
Mass of excess Li =
Explanation of Solution
2 moles of Li will form 2 moles of LiBr
Thus, 0.312 mole of LiBr was formed from 0.312 mole of Li.
Also, 3.60 mole of lithium is present.
From that amount 0.312 mole reacts, hence remaining Li = 3.60 − 0.312 = 3.288 mole
3.288 mole of Li is in excess
Mass of excess Li =
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