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3 years ago

Which equations represent precipitation reactions?

Chemistry

1 answer:

viktelen [127]3 years ago

3 0

In a precipitation reaction, one of the product formed is an insoluble precipitate.

The given reactions can be written with phases included as:

A. $Na_{2}S(aq)+FeBr_{2}(aq)-->2NaBr(aq)+FeS(s)$ : In this reaction one of the product, FeS is insoluble. Therefore, this is a precipitation reaction.

B. $MgSO_{4}(aq)+CaCl_{2}(aq)-->MgCl_{2}(aq)+CaSO_{4}(s)$ : In this reaction, the product $CaSO_{4}$ is a solid(insoluble). So, this is a precipitation reaction too.

C. $LiOH(aq)+NH_{4}I(aq)-->LiI(aq)+NH_{4}OH(aq)$ : In this reaction, both the products are soluble. So this is not a precipitation reaction.

D. $2NaCl(aq)+K_{2}S(aq)-->Na_{2}S(aq)+2KCl(aq)$ : In this reaction, both the products are soluble. So this is not a precipitation reaction.

E. $AgNO_{3}(aq)+NaCl(aq)-->AgCl(s)+NaNO_{3}(aq)$ : In this reaction, the product AgCl is a precipitate. So, it is a precipitation reaction.

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Answer:

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

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