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maksim [4K]
3 years ago
15

Balance the following equation; then determine how many grams of CO2 will be produced when 40.0 mol of CO reacts with Fe2O3.

Chemistry
1 answer:
sashaice [31]3 years ago
6 0

 Fe2O3      +       3CO -------->    2Fe    +      CO2

      1           :          3         :            2      :         3

   13.3       <---     40    ------> 26.6      --->     40    ( mol)

n = m/M

m CO2 = n.M = 13.3  .    40   =  532 ( g)

p/s : i hope that this will help  ( cause i'm not really good at english :}}} )

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Answer:

The answer is Dissemination of components controlled by electron design.  

Explanation:

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7 0
3 years ago
The titration of 25.0 mL of an iron(II) solution required 18.0 mL of a 0.145 M solution of dichromate to reach the equivalence p
Svetlanka [38]

Answer:

0.64 M

Explanation:

Given:

Volume of iron(II) solution (V₁) = 25.0 mL = 0.025 L

Molarity of iron(II) solution (M₁) = ?

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Volume of dichromate solution (V₂) = 18.0 mL = 0.018 L

Molarity of dichromate solution (M₂) = 0.145 M

Number of moles of dichromate solution (n₂) = ?

Molarity is equal to the ratio of moles and volume.

So, molarity of dichromate solution is given as:

M_2=\frac{n_2}{V_2}\\\\n_2=M_2\times V_2=0.145\times 0.018 = 2.61\times 10^{-3}\ mol

Now, let us write the complete balanced reaction for the given situation.

So, the complete balanced equation is given below.

6Fe^{2+}(aq)+Cr_2O_7^{2-}(aq)+14H^+(aq)\to 6Fe^{3+}(aq)+2Cr^{3+}(aq)+7H_2O

From the equation, it is clear that, 1 mole of dichromate is required for 6 moles of iron(II) solution.

So, using unitary method, we find the number of moles of iron(II) solution.

1 mole of dichromate = 6 moles of iron(II)

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So, 0.016 moles of iron(II) is needed. Therefore, n_1=0.016\ mol

Now, molarity of iron(II) solution is given as:

Molarity = Moles ÷ Volume

M_1=\frac{n_1}{V_1}\\\\M_1=\frac{0.016\ mol}{0.025\ L}=0.64\ M

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4 0
3 years ago
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Hello!

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Best regards!

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