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3 years ago

Following reactionsFeSO4+KOH

Chemistry

1 answer:

prohojiy [21]3 years ago

5 0

Explanation:

$FeSO _{4} + \: KOH \: →K _{2}SO _{4} \: + \: Fe(OH) _{2}$

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Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a

horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound Amount

Ni 12.7 g

CO 1.98 g

$Ni(CO)_4$ 0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of $Ni(CO)_4$ = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

$\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M$

$\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M$

$\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M$

For the given chemical reaction:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

The expression of equilibrium constant for the reaction:

$K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}$

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1$

Hence, the value of equilibrium constant for the reaction is 2448.1

4 0

4 years ago

Question 5

Zina [86]

Answer:

Explanation:

2.2 g/mL is the answer.

4 0

3 years ago

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc

docker41 [41]

Answer:

The theoretical yield of NH3 produced by the reaction is 16.89 grams

Explanation:

Step 1: Data given

Mass of N2 = 13.90 grams

Molar mass of N2 = 28.0 g/mol

Mass H2 = 6.440 grams

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

N2(g) + 3 H2(g) → 2 NH3(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2 = 13.90 grams / 28.0 g/mol

Moles N2 = 0.496 moles

Moles = 6.440 grams / 2.02 g/mol

Moles H2 = 3.19 moles

Step 4: Calculate limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.496 moles)

H2 is in excess. There will react 3*0.496 = 1.488 moles

there will remain 3.19 - 1.488 = 1.702 moles

Step 5: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.496 moles N2 we will have 2*0.496 = 0.992 moles NH3

Step 6: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.992 moles * 17.03 g/mol

Mass NH3 = 16.89 grams

The theoretical yield of NH3 produced by the reaction is 16.89 grams

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3 years ago

How many grams of Pb and FeSO4 are required to produce 200 grams of PbSO4?, what mass of Fe is also produced?

Akimi4 [234]

Explanation:

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3 years ago

Earths surface is warmed by heat, ___ by the sun.

zavuch27 [327]

Radiated would be the answer

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3 years ago

Following reactionsFeSO4+KOH​

Following reactionsFeSO4+KOH