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2 years ago

How many grams of oxygen are formed when 8.20 moles of KOH are formed? 4 KO(s) + 2 H2O(l) → 4 KOH(s) + O2(g)

Chemistry

2 answers:

julia-pushkina [17]2 years ago

6 0

Answer : The mass of oxygen formed are 65.6 grams.

Explanation :

First we have to calculate the moles of $O_2$

The given balance chemical reaction is:

$4KO(s)+2H_2O(l)rightarrow 4KOH(s)+O_2(g)$

From the balanced chemical reaction we conclude that,

As, when 4 moles of KOH formed then 1 mole of $O_2$ will be formed

So, when 8.20 moles of KOH formed then $\frac{8.20}{4}=2.05$ mole of $O_2$ will be formed

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

Molar mass of $O_2$ = 32 g/mole

$\text{ Mass of }O_2=(2.05moles)\times (32g/mole)=65.6g$

Therefore, the mass of oxygen formed are 65.6 grams.

Pie2 years ago

3 0

49.7g02* that's the answer

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1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

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$CH_{3}COOH(aq)$ act as Acid

$NaHCO_{3}(aq)$ act as weak base

Reactant : $CH_{3}COOH(aq)$ , $NaHCO_{3}$

Number of atoms of :

C = 2 ( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 2 + 1

= 3

H = 4( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 4 + 1

O = 2( $CH_{3}COOH(aq)$ ) + 3 ( $NaHCO_{3}$ )

= 5

Na = 1 ( $NaHCO_{3}$ )

= 1

Product : $CO_{2}(g)$ , $H_{2}O(l)$ , $CH_{3}COONa(aq)$

Number of atoms :

C = 1( $CO_{2}(g)$ ) + 2( $CH_{3}COONa(aq)$ )

= 1 + 2

= 3

H = 2( $H_{2}O(l)$ ) + 3( $CH_{3}COONa(aq)$ )

= 2 + 3

= 5

O = 1( $H_{2}O(l)$ ) + 2( $CH_{3}COONa(aq)$ )

+2( $CO_{2}(g)$

= 1 + 2 + 2

= 5

Na = 1( $CH_{3}COONa(aq)$

= 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

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Check the balancing in both reactant and products should be :

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H = 2

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