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Alecsey [184]
3 years ago
9

Iron(II) sulfide reacts with hydrochloric acid according to the reaction:

Chemistry
2 answers:
Inessa [10]3 years ago
8 0
The balanced chemical reaction would be:

FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g) 

We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:

Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl

0.646 - 0.48 = 0.166 mol HCl left
zysi [14]3 years ago
4 0

Answer: The amount (in moles) of the excess reactant left is, 0.166 mol

Explanation : Given,

Moles of FeS = 0.240 m ol

Moles of HCl = 0.646 mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)

From the balanced reaction we conclude that

As, 1 mole of FeS react with 2 mole of HCl

So, 0.240 moles of FeS react with 0.240\times 2=0.48 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and FeS is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of HCl

Remaining moles of HCl = 0.646 - 0.48 = 0.166 mol

Thus, the amount (in moles) of the excess reactant left is, 0.166 mol

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What steps should you take to respond to an accident?
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Which describes an impermeable rock where oil and natural gas deposits are found?
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Answer:

Explanation:

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5 0
3 years ago
How many kilograms of solvent (water) must 0.71 moles of KI be dissolved in to produce a 1.93 m solution?
GalinKa [24]

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6 0
4 years ago
PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC
Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

                PCl₃(g)        +        Cl₂(g)     ⇋        PCl₅(g)

Initially     0.24 M                 1.50M                 0.12 M

React           x                           x                         x

Some amount of compound has reacted during the process.

In equilibrium we have

              0.24 - x                  1.50 - x                  0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)          

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0
4 years ago
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