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mart [117]
2 years ago
12

Please help! 10 points

Chemistry
1 answer:
myrzilka [38]2 years ago
4 0

Answer:

<u>1 mole of Fe and 1.5 moles of CO2</u>

Explanation:

<u>Fe2O3 + CO → Fe + CO2</u>

This is a trick question.  The equation is not balanced properly.  To start, note that one molecule of Fe2O3 contain 2 Fe atoms, while there is only 1 Fe atom on the product side.  And the oxygens are also missing (from 4 to 2).  We cannot use this equation, as written.

Here's one that is balanced:

<h2><u>Fe2CO3 + 3CO = 2Fe + 3CO2</u></h2>

<u>I'll use this one to answer the question.</u>

1.  <u>If 3 moles of Fe2O3 react with 1.5 moles of CO, how many moles of each product are formed?</u>

The balanced equation tells us that 1 mole of Fe2O3  will react with 3 moles CO to produce 2 moles of Fe and 3 moles of CO2.  <u><em>If we had enough CO</em></u>,  3 moles Fe2O3 will produce 6 moles of Fe and 9 moles of CO2.  But note that we'd also need 9 moles of CO for a complete reaction.  We only have 1.5 moles, probably due to Stanley, our new manager.  So we're limited by the CO, and we'll have to delete our first analysis and work from 1.5 moles of CO.

<h2>If we have 1.5 moles CO, we'll produce:</h2><h2> <u>1 mole of Fe and 1.5 moles CO2.</u></h2>
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3 years ago
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

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3 years ago
A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
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Answer:

The mass of water m_{w} = 39.18 gm

Explanation:

Mass of iron m_{iron} = 32.5 gm

Initial temperature of iron T_{1} = 22.4°c = 295.4 K

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Mass of water = m_{w}

Specific heat of water  C_{w} = 4.2 \frac{KJ}{kg  K}

Initial temperature of water T_{2} = 336 K  

Final temperature after equilibrium T_{f} = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water  = Heat gain by iron rod

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Put all the values in above formula we get

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Answer:

6.31g/mol

Explanation:

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* Mm = m/n

Also, density (p) = mass (m) ÷ volume (V)

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PV = M/Mm. RT

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Since p = m/V

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