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2 years ago

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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.

1 answer:

Sladkaya [172]2 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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What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH

AlladinOne [14]

1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol

</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO

<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.

CO + 2H2 -------> CH3OH
   1 mol   2 mol
given 2.66 mol   6 mol (excess)

How much
we need CO?    3 mol 6 mol

We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.

3)   </span>CO    + 2H2 ------->   CH3OH
   1 mol    1 mol
  2.66 mol 2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 +  4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH =  85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.

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3 years ago

Definition of contourology

agasfer [191]

Science of, or study of contours, or the lines outlining a scene, drawing, etc.

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2 years ago

The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?

svlad2 [7]

Answer: Dissociation constant of the acid is $10^{-11}$ .

Explanation: Assuming the acid to be monoprotic, the reaction follows:

$HA\rightleftharpoons H^++A^-$

pH of the solution = 6

and we know that

$pH=-log([H^+])$

$[H^+]=antilog(-pH)$

$[H^+]=antilog(-6)=10^{-6}M$

As HA ionizes into its ions in 1 : 1 ratio, hence

$[H^+]=[A^-]=10^{-6}M$

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

$[HA]=[HA]-[H^+]$

$[HA]=0.1M-10^{-6}M$

$[HA]=0.09999M$

Dissociation Constant of acid, $K_a$ is given as:

$K_a=\frac{[A^-][H^+]}{HA}$

Putting values of $[H^+],[A^-]\text{ and }[HA]$ in the above equation, we get

$K_a=\frac{(10^{-6}M).(10^{-6}M)}{0.09999M}$

$K_a=1.0001\times 10^{-11}M$

Rounding it of to one significant figure, we get

$K_a=1.0\times 10^{-11}M\approx 10^{-11}M$

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2 years ago

formic acid, hcooh, is a weak acid with a ka equal to 1.8×10–4. what is the ph of a 0.0115 m aqueous formic acid solution?

kvasek [131]

The pH of a 0.0115 m aqueous formic acid solution is mathematically given as

pH=2.8424

This is further explained below.

<h3>What is the ph of a 0.0115 m aqueous formic acid solution?</h3>

Generally, the equation for the chemical equation is mathematically given as

HCOOH H^+ + HCOO

$\quad C-C \alpha \quad C \alpha$

$&k a=\frac{c^2 \alpha^2}{e(1-\alpha)}\\\\&K_a=\frac{c \alpha^2}{1-\alpha_4} \quad|\gg\rangle \alpha\\\\&K_a=c \alpha^2\\\\&\alpha=\sqrt{\frac{k a}{c}}\\\\&\alpha=\sqrt{\frac{1.8 \times 10^{-4}}{0.0115}}\\\\&\alpha=\sqrt{156: 527710^{-4}}\\\\&\alpha=\sqrt{1.565 \times 1 \sigma^2}\\\\&\alpha=1.25 \times 10^{-1}\\\\&\alpha=0.125\\$

$&\left[\mathrm{H}^{+}\right]=\mathrm{C \alpha}\\\\&=0.125 \times 0.0115\\\\&=1.4375 \times 10^{-9}\\\\&P=-\log \left[H^{+}\right]\\\\&=-\log \left[1.4375 \times 10^{-3}\right]\\\\&P H=2.8424$

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