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Alex
3 years ago
6

How many grams of Fe are produced from 65.9 g of FeO given the following reaction: 3FeO + 2Al -> 3Fe +Al2O3

Chemistry
2 answers:
patriot [66]3 years ago
8 0

Using the mole ratio and stoichiomrtery, you get:

65.9 g FeO x (molar mass of FeO) x (3 mol Fe/ 3 mol FeO) x (molar mass of Fe).

Plug in the values and you’re good to go!

klasskru [66]3 years ago
4 0
65.9 g FeO divided by 72 ( molar mass of feo) then divided by 3 multiplied by 3 multiplied by 56 =>
About 54 Grams
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Sophie [7]
A is correct because there is no reaction involved
6 0
3 years ago
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Determine the enthalpy change when 23.5 g of carbon is reacted with oxygen according to the reaction C(s) + O2(g) CO2(g) delta H
xeze [42]
We are provided with the amount of energy released when one mole of carbon reacts. We mus first convert the given mass of carbon to moles and then compute the energy released for the given amount.

Moles = mass / atomic mass
Moles = 23.5 / 12
Moles = 1.96 moles


One mole releases 394 kJ/mol
1.96 moles will release:
394*1.96
= 772.24

The enthalpy change of the reaction will be -772.24 kJ
8 0
2 years ago
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)
katen-ka-za [31]

Answer:

3.81 g Pb

Explanation:

When a lead acid car battery is recharged, the following half-reactions take place:

Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)  

Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻

We can establish the following relations:

  • 1 A = 1 c/s
  • 1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
  • The molar mass of Pb is 207.2 g/mol
  • 1 mol of e⁻ has a charge of 96468 c (Faraday's constant)

Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is:

37.0s.\frac{96.0c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molPb}{2mole^{-} } .\frac{207.2gPb}{1molPb} =3.81gPb

6 0
3 years ago
Match these items with their examples.
kiruha [24]
Here you go! Feel free to ask questions!

7 0
3 years ago
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A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.
lord [1]

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

(31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g

The produced energy will be:

=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K

=450.35\times 3.2

=1441.12 \ J

The reaction will be:

⇒  HCl+NaOH \rightarrow NaCl+H_{2}O

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.  

Therefore, the moles generated by NaCl will indeed be:

=  (\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}

=  0.0318\times 0.500

=  0.0159 \ mole  \ of \ NaCl

Now,

=  \frac{1441.12 \ J}{0.0159 \ moles \ NaCl}

=  906364.7

=  90.6 \ KJ/mol \ NaCl

7 0
2 years ago
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