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2 years ago

PLEASE HELP

Chemistry

2 answers:

Marta_Voda [28]2 years ago

5 0

The answer for this is D

OverLord2011 [107]2 years ago

5 0

The answer is D purr:)

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What are the products of a neutralization reaction?

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In an acid and base neutralization reaction, the products are salt and water.

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Which of the following statements is true of enzymes A. They're unusable after a reaction B. They don't provide an active site f

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D. They act on a specific type of substrate in a reaction

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3 years ago

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Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate

Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of $K_3PO_4$ = 150 g

Mass of $Al_2(CO_3)_3$ = 90 g

Molar mass of $K_3PO_4$ = 212.27 g/mole

Molar mass of $Al_2(CO_3)_3$ = 233.99 g/mole

Molar mass of $K_2CO_3$ = 138.205 g/mole

First we have to calculate the moles of $K_3PO_4$ and $Al_2(CO_3)_3$

$\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles$

$\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles$

The given balanced reaction is,

$2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4$

From the given reaction, we conclude that

2 moles of $K_3PO_4$ react with 1 mole of $Al_2(CO_3)_3$

0.7066 moles of $K_3PO_4$ react with $\frac{1}{2}\times 0.7066=0.3533$ moles of $Al_2(CO_3)_3$

But the moles of $Al_2(CO_3)_3$ is, 0.3846 moles.

So, $Al_2(CO_3)_3$ is an excess reagent and $K_3PO_4$ is a limiting reagent.

Now we have to calculate the moles of $K_2CO_3$ .

As, 2 moles of $K_3PO_4$ react to give 3 moles of $K_2CO_3$

So, 0.7066 moles of $K_3PO_4$ react to give $\frac{3}{2}\times 0.7066=1.0599$ moles of $K_2CO_3$

Now we have to calculate the mass of $K_2CO_3$ .

$\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3$

$\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g$

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

$\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100$

$\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%$

Therefore, the % yield of potassium carbonate is, 85.33%

6 0

2 years ago

Fe+CuCI2> FeCI2+Cu balancing equations

Volgvan

Explanation:

To balance a chemical equation implies that we are conserving the number and mass of the reacting species.

Chemical equations must obey the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed".

This suggests that mass is conserved.

The number of atoms on the reactant side must be equal to that on the product end.

Before proceeding, check the species to see if they have equal number of atoms on both sides:

Fe + CuCI₂ → FeCI₂ + Cu

Elements reactants products

Fe 1 1

Cu 1 1

Cl 2 2

We can see that the equation is already balanced

learn more:

balanced equation brainly.com/question/2612756

#learnwithbrainly

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