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Vladimir [108]
3 years ago
14

In 1901, Thomas Edison invented the nickel-iron battery. The following reaction takes place in the battery. Fe(s) + 2 NiO(OH)(s)

+ 2 H2O(l) Fe(OH)2(s) + 2 Ni(OH)2(aq) How many mole of Fe(OH)2, is produced when 5.35 mol Fe and 7.65 mol NiO(OH) react?
Chemistry
1 answer:
AveGali [126]3 years ago
5 0

Ans: Moles of Fe(OH)2 produced is 5.35 moles.

Given reaction:

Fe(s) + 2NiO(OH) (s) + 2H2O(l) → Fe(OH)2(s) + 2Ni(OH)2(aq)

Based on the reaction stoichiometry:

1 mole of Fe reacts with 2 moles of NiO(OH) to produce 1 mole of Fe(OH)2

It is given that there are:

5.35 moles of Fe

7.65 moles of NiO(OH)

Here the limiting reagent is Fe

Therefore, number of moles of Fe(OH)2 produced is 5.35 moles.






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How many grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron? Fe2O3(s
erastova [34]

Answer : The grams of carbon monoxide needed are 148.89 g

Solution : Given,

Mass of iron, Fe = 198.5 g

Molar mass of iron, Fe = 56 g/mole

Molar mass of carbon monoxide, CO = 28 g/mole

First we have to calculate the moles of iron, Fe.

\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{198.5g}{56g/mole}=3.545moles

The balanced chemical reaction is,

Fe_2O_3(s)+3CO(g)\rightarrow 3CO_2(g)+2Fe(s)

From the balanced reaction, we conclude that

2 moles of iron produces from the 3 moles of carbon monoxide

3.545 moles of iron produces from the \frac{3}{2}\times 3.545=5.3175 moles of carbon monoxide

Now we have to calculate the mass of carbon monoxide, CO.

\text{ Mass of CO}=\text{ Moles of CO}\times \text{ Molar mass of CO}

\text{ Mass of CO}=(5.3175moles)\times (28g/mole)=148.89g

Therefore, the grams of carbon monoxide needed are 148.89 g

7 0
2 years ago
For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+
Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

    H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
3 years ago
For the reaction below, if the rate of appearance of Br2 is 0.180 M/s, what is the rate of disappearance of
brilliants [131]

Answer:

–0.360 M/s

Explanation:

7 0
3 years ago
You are using a ruler to measure the width of a test tube. The edge of the test tube is between 3 cm and 4 cm marks. if centimet
Arada [10]

Answer:

A) 3.6 cm

Explanation:

Accuracy comes down to how precisely you can read the length on a given scale. Here since the smallest increment is centimeter, we can go only one decimal beyond to estimate. This is because you can usually estimate to only one decimal place beyond the closest marks on any measuring.

So, the answer should be 3.6 cm.

Here's a document that explains it well: https://www.auburn.wednet.edu/cms/lib03/WA01001938/Centricity/Domain/1360/1_Uncertainty.pdf

Hope that's right!

5 0
1 year ago
In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i
Alborosie
<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

  • When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
  • From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
  • For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
  • Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.

3 0
3 years ago
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