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Fiesta28 [93]
4 years ago
9

A water bottling company is looking to achieve maximum production from the following reversible reaction. If the reaction has re

ached a point of equilibrium, what would be the best way to maximize production of water? 2 H2+O2⇔2 H2O Remove hydrogen and allow the reaction to reach equilibrium again. Stop of the process because no more water can be made after equilibrium is reached. Remove the already produced water and allow the reaction to reach equilibrium again. Remove oxygen and allow the reaction to reach equilibrium again.
Chemistry
1 answer:
dybincka [34]4 years ago
7 0

Answer:

Remove the already produced water and allow the reaction to reach equilibrium again.

Explanation:

<em>According to Le Chatelier principle, when a reaction is in equilibrium and one of the factors that influence the rate of reaction is altered, the equilibrium will shift so as to annul the effects of the change.</em>

If the product is continuously being removed from a reaction that is in equilibrium, more product will continued to be formed in another to annul the effect of reduction in the concentration of product.

Hence, in order to maximize production of water in the reaction, the product (water) needs to be removed and the reaction allowed to reach equilibrium again.

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A

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givi [52]

Answer:

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Explanation:

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6 0
3 years ago
What mass of lead (II) chloride is produced when 200.0 mL of a 0.250 M solution of sodium chloride is mixed with 200.0 mL of a 0
kow [346]

Answer:

Option d. 6.95 g

Explanation:

First of all, we state the reaction:

2NaCl + Pb(NO₃)₂ → PbCl₂ +  2NaNO₃

We determine the moles of each reactant, to state the limiting

Firstly we convert volume frm mL to L

0.200 L . 0.250M = 0.05 moles of NaCl

0.200L . 0.250M = 0.05 moles of Pb(NO₃)₂

Acording to stoichiometry we know that relation is 1:2, so the limiting reagent is the NaCl.

For 1 mol of Pb(NO₃)₂ I need 2 moles of NaCl

For 0.05 moles of Pb(NO₃)₂ I would need, the double → 0.1 moles

(We only have, 0.05 moles of NaCl)

Stoichiometry to the formed product is 2:1

From 2 moles of NaCl I produce 1 mol of PbCl₂

From 0.05 moles I would produce, the half → 0.025 moles

Let's convert the moles to mass → 0.025 mol . 278.1 g / 1mol = 6.95 g

8 0
3 years ago
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