Chemistry, 14.12.2020 17:10, bjpvrpow74wq

You need 150g of pure lithium for an experiment you're doing. You have 675g of lithium oxide (Li2O). Can you extract all the lithium you need from the amount of compound you have? Show your reasoning.

Answers: 2

Show answers

Answers

answered: maddie1776

Yes, you can extract all the lithium you need from the compound

We'll begin by calculating the mass of lithium, Li in 1 mole of Li₂O. This can be obtained as described below:

1 mole of Li₂O = (2×7) + 16

1 mole of Li₂O = 14 + 16 = 30 g

SUMMARY

30 g of Li₂O contains 14 g of Li.

With the above information in mind, we can determine the mass of Li in 675 g of lithium oxide, Li₂O. This can be obtained as follow:

30 g of Li₂O contains 14 g of Li.

Therefore,

675 g of Li₂O Will contain = (675 × 14)/30 = 315 g of Li.

From the calculation made above, we can see that there are 315 g of Li in 675 g of Li₂O. Thus, we can conclude that 150 g of Li can be extracted from the compound (i.e 675 g of Li₂O)

Learn more: link

answered: lucifer6669

Yes we can extract all the lithium from the amount of compound we have.

Explanation:

To calculate the moles :

$moles=\frac{\text {given mass}}{\text {Molar mass}}$

$\text{Moles of} Li=\frac{150g}{7g/mol}=21.4moles$

$\text{Moles of} Li_2O=\frac{675g}{30g/mol}=22.5moles$

$2Li_2O\rightarrow 4Li+O_2$

According to stoichiometry :

2 moles of Li_2O produce 4 moles of

Thus 22.5 moles of Li_2O will produce= $\frac{4}{2}\times 22.5=45moles$ of Li

As 21.4 moles is a lesser quantity than 45 moles, thus it can be produced

answered: Guest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answered: Guest

Qand k are only the same when the forward rate of the reaction has reached zero.