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3 years ago

Which solution is the most concentrated

Chemistry

1 answer:

san4es73 [151]3 years ago

5 0

Answer:

The most concentrated solution is the one with the highest molarity. Molarity is calculated by taking the moles of solute/ liters of solution. The already gave you the molarity, so you do not need to calculate it. The most concentrated solution is the 0.25 M KCl.

Hope this helps!

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Explanation:

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Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g) When 63.1 g of HCl is allowed to react with 17.2 g of O2

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Answer:

The limiting reactant for this reaction is the HCl

Explanation:

This is my reaction:

4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)

Molar mass O2 = 32 g/mol

Molar mass HCl = 36,45 g/mol

Mass / Molar mass = Moles

Moles HCl : 63,1 g / 36,45 g/m = 1,73 moles

Moles O2: 17,2 g / 32g/m = 0,54 moles

4 moles of HCl react with 1 mol of O2, according to reaction, so

1,73 moles of HCl, are going to react with, how many moles of O2.

4 moles HCl ___ 1 mol O2

1,73 moles HCl ___ (1,73 . 1)/ 4 = 0,43 moles of O2

O2 is my excess reagent because I need 0,43 moles and I have 0,54 so I have moles in excess.

1 mol of O2 are going to react with 4 moles of HCl

0,54 moles of O2 are going to react with, how many moles of HCl ?

1 mol O2 ____ 4 moles HCl

0,54 mol O2 ___ (0,54 . 4)/ 1 = 2,16 moles

I need 2,16 moles to consume my moles of O2, but I only have 1,73 moles, tha's why the HCl is mi limiting reactant.

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Read 2 more answers

Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

kumpel [21]

Answer : The limiting reactant is $O_2$

Explanation : Given,

Mass of $C_3H_8$ = 30.0 g

Mass of $O_2$ = 75.0 g

Molar mass of $C_3H_8$ = 44 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_3H_8$ and $O_2$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{30.0g}{44g/mole}=0.682moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{75.0g}{32g/mole}=2.34moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction we conclude that

As, 5 mole of $O_2$ react with 1 mole of $C_3H_8$

So, 2.34 moles of $O_2$ react with $\frac{2.34}{5}\times 1=0.468$ moles of $C_3H_8$

From this we conclude that, $C_3H_8$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Therefore, the limiting reactant is $O_2$

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3 years ago