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Which of the following statements is//ar...

Which of the following statements `is//are` correct?
i. 21.0 o lithium reacts with `32.0 g "of"O_(2)`.
`4 Li + O_(2) rarr 2Li_(2) O`
ii. `3.9 g` of K reacts with `4.26 g "of" Cl_(2)`
`2K + Cl_(2) rarr 2 KCl`
Atomic weights of `Li = 7` and `K = 39`. `Mw "of" Li_(2) O = 30` and `KCl = 74.5 g mol^(-1)`

A

In reaction (i), `O_(2)` is in excess.

B

`45.0 g of Li_(2)O` is formed in reaction (i)

C

In reaction (ii), `Cl_(2)` is in excess.

D

`7.45 g of KCl` is formed is reaction (ii).

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Moles of `Li = (21)/(7) = 3.0 "mol"`
Moles of `O_(2) = (32)/(32) = 3.0 "mol"`
a. Since `(3.0 "mol" Li) ((1 "mol" O_(2))/(4 "mol" Li)) = (3)/(4) = 0.075 "mol of" O_(2)`is required, therefore `(1.0 - 0.75) = 0.25 "mol of" O_(2)` is
in excess. Hence, `Li` is the limiting reagent
Weight of `Li_(2) O` formed
`(3.0 "mol" Li) ((1 "mol" Li_(2) O)/(2 "mol" Li))((30 g Li_(2) O)/("mol" Li_(2)O))`
`= (3 xx 1 xx 30)/(2) = 45.0 g Li_(2)O`
c. Moles of `K = (3.9)/(39) = 0.1 "mol"`
Moles of `Cl_(2) = (4.26)/(71) = 0.06 "mol"`
Since `((0.1)/(2)) = 0.05` mol of `Cl_(2)` is required. Therefore, `(0.06 - 0.05) = 0.01` mol of `Cl_(2)` is in excess.
Hence `K` is the limiting reagent
Weight of `KCl` formed
`= (0.1 "mol" K) ((1 "mol" KCl)/("mol" K))((74.5 g KCl)/("mol"KCl"))`
`= 0.1 xx 1 xx 74.5 = 7.45 g "of" KCl`
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