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2 years ago

1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →

Chemistry

1 answer:

Leno4ka [110]2 years ago

5 0

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

<h3>What is Limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

As 80 g NaOH produces 106 g of Na₂CO₃.

Therefore 5 g NaoH will produce ;

106 / 80 x 5 = 6.625 g

Learn more about limiting reagent here ;

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Answer:

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Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{K_m+[S]}=k_{cat}[E_o]\times \frac{[S]}{K_m+[S]}$

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We have :

$K_m=0.045 mol/dm^3$

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$[S]=0.110 mol/dm^3$

$v=V_{max}\times \frac{[S]}{K_m+[S]}$

$1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}$

$V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s$

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

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3 years ago

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3 years ago

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USPshnik [31]

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The equation for the reaction would be;

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3 years ago

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