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BigorU [14]
3 years ago
14

You need 150g of pure lithium for an experiment you're doing. You have 675g of lithium oxide (Li2O). Can you extract all the lit

hium you need from the amount of compound you have? Show your reasoning.
Chemistry
1 answer:
Sladkaya [172]3 years ago
7 0

Answer: 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 Li

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

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Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9 at 1,000 K

Calculate Kc for the reaction 2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

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The given chemical equations follows:

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c

We need to calculate the equilibrium constant for the equation, which is:

2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

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K_c'=(K_c)^2

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K_c=4.4\times 10^9

Putting values in above equation, we get:

K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}

Hence, the value of K_c' for the final reaction is 1.936\times 10^{19}

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