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Alecsey [184]
2 years ago
9

For the reaction of hydrogen with iodine

Chemistry
1 answer:
fenix001 [56]2 years ago
8 0

Answer:

r_{H_2} = \frac{-1}{2} r_{HI}

Explanation:

Hello!

In this case, considering the given chemical reaction:

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

Thus, by applying the law of rate proportions, we can write:

\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

r_{H_2} = \frac{-1}{2} r_{HI}

Best regards!

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2.56 g of hydrogen reacts completely with 20.32 g of oxygen<br> to form X g of water. X = g
Brilliant_brown [7]

Answer:

Mass of water produced is 22.86 g.

Explanation:

Given data:

Mass of hydrogen = 2.56 g

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Solution:

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2H₂ + O₂   →  2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 20.32 g/ 32 g/mol

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Number of moles = mass/ molar mass

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Number of moles = 1.28 mol

Now we will compare the moles of water with oxygen and hydrogen.

                    O₂            :            H₂O

                     1              :             2

                  0.635        ;            2×0.635 =  1.27

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                    2              :              2

                 1.28            :           1.28

The number of  moles of water produced by oxygen are less thus it will be limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 1.27 × 18 g/mol

Mass = 22.86 g

 

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