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Usimov [2.4K]
3 years ago
7

The overall reaction in a commercial heat pack can be represented as How much heat is released when 4.40 moles of iron are react

ed with excess ? Heat = kJ How much heat is released when 1.00 mole of is produced? Heat = kJ How much heat is released when 1.60 g iron is reacted with excess ? Heat = kJ How much heat is released when 11.8 g and 1.20 g are reacted? Heat = kJ
Chemistry
1 answer:
vodomira [7]3 years ago
7 0

Answer:

Explanation:

The overall equation for this reaction can be represented as:4 Fe(s) + 3 O_{2(g) }  \to 2Fe_2O_{3(s)} \ \ \ \      \Delta H = -1652  \ kJ

The first question says:

How much heat is released when 4.40 moles of iron is reacted with excess O₂?

Suppose 1652 kJ of heat is being emitted into the surroundings when four(4) moles of Fe reacted with O₂, therefore;

4.40 moles of Fe reacts with:

=\dfrac{4.40 \ moles \times  1652 \ kJ}{4 \ moles}

= 1817.2 kJ of heat will be produced.

The second question says:

How much heat is released when 1.00 mole of Fe_2O_3 is produced?

Given that 1652 kJ of heat is being emitted into the surroundings when two(2) moles of Fe_2O_3 is produced, therefore;

1.00 moles of Fe_2O_3 reacts with:

=\dfrac{1.00 \ moles \times  1652 \ kJ}{2 \ moles}

= 826 kJ of heat will be produced.

To the third question; we have:

How much heat is released when 1.60 g iron is reacted with excess O₂?

We need to find the number of moles of iron first.

We know that number of moles = mass/molar mass

Thus, the molar mass of iron = 55.8 g/mol

number of moles of iron = (1.60g) / (55.8 g/mol)

number of moles of iron = 0.02867 mol

Thus; \dfrac{0.02867\  mol \times  1652 \ kJ }{4 \ mol}

= 11.84 kJ of heat is released.

The last question says:

How much heat is released when 11.8 g Fe and 1.20 g O₂ are reacted?

Again;

the number of moles of Fe = (11.8g) / (55.8 g/mol) = 0.2114 mole of Fe

Thus; \dfrac{0.2114\  mol \times  1652 \ kJ }{4 \ mol}

= 87.31 kJ of heat is released.

On the other hand,

the number of moles of O₂ = (1.20g) / (32 g/mol) = 0.0375 mol of O₂

Thus; \dfrac{0.0375\  mol \times  1652 \ kJ }{3 \ mol}

= 20.65 kJ of heat is released

Therefore, when these two(2) reactants reacted with each other, it is just the smaller amount of heat that would be released because oxygen tends to be the limiting reactant.

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Answer:

Water molecules feature the combinations of hydrogen and oxygen atoms in a 2:1 ratio. Since they are present in a fixed ratio of mass, water molecules obey the law of constant proportions. Water is formed when two molecules of the diatomic hydrogen gas, combine with one molecule of the diatomic oxygen gas to produce two molecules of water

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A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3
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Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of Pb(IO_{3})_{2} is given as follows-

                   Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}

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                             [Pb^{2+}]=S(M) and [IO_{3}^{-}]=2S(M)

Where species under third bracket represent equilibrium concentrations

So, solubility product of Pb(IO_{3})_{2} , K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}

Here, [Pb^{2+}]=S(M)=5.0\times 10^{-5}M

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Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c
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Answer:

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}  \\

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

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