A reaction involves the following elementary steps:

Chemistry

1 answer:

olga55 [171]4 years ago

7 0

Answer:

Explanation:

H₂ $_{g}$ + O₂ $_{g}$ → H₂O₂ $_{g}$

H₂ $_{g}$ + H₂O₂ $_{g}$ → 2H₂O $_{g}$

The sum of the two equations is a combined equation where all the compounds on the reactant sides are written on the Left hand side and the products are all on the right hand side:

H₂ $_{g}$ + O₂ $_{g}$ + H₂ $_{g}$ + H₂O₂ $_{g}$ → H₂O₂ $_{g}$ + 2H₂O $_{g}$

The net reaction equation involves the cancellation of species that occurs on both sides, these species are H₂O₂:

We then have:

H₂ $_{g}$ + H₂ $_{g}$ + O₂ $_{g}$ → 2H₂O $_{g}$

combining the two repeating hydrogen gas:

2H₂ $_{g}$ + O₂ $_{g}$ → 2H₂O $_{g}$

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What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)

astra-53 [7]

Answer:-

Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.

For KOH, $10.0mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})$

= 0.002 mol

For Cinnamic acid, $30.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})$

= 0.003 mol

Acid and base react as:

$HC_9H_7O_2(aq)+KOH(aq)\rightleftharpoons KC_9H_7O_2(aq)+H_2O(l)$

The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.

0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001

As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$