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3 years ago

How many grams of LiNO2 are required to make 250 mL of 0.75 Molarity solution?

Chemistry

1 answer:

Katarina [22]3 years ago

8 0

Answer:

159 g OF LiNO2 WILL BE USED TO MAKE 0.25 L OF 0.75 M SOLUTION.

Explanation:

How many grams of LiNO2 are required to make 250 mL of 0.75 M?

First calculate the molarity in mol per dm3

0.75 M of LiNO2 reacts in 250 mL = 250 /1000 L volume

0.75 M = 0.25 L

In 1 L, the molarity of LiNO2 will be:

= (0.75 * 1/ 0.25) M

= 3 mol/dm3 of LiNO2

Next is to calculate the molarity in g/dm3:

Molarity in mol/dm3 = molarity in g/dm3 / RMM.

RMM of LiNO2

(Li = 7 , N =14 , 0 = 16)

RMM = ( 7 + 14 + 16 * 2) = 53 g/mol

Molarity in mol/dm3 = Molarity in g/dm3 / RMM

Molarity in g/dm3 = Molarity in mol/dm3 * RMM

Molarity in g/dm3 = 3 * 53

Molarity in g/dm3 = 159 g/dm3.

So therefore, to make 250 mL of 0.75 M of LiNO2, we use 159 grams of LiO2.

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(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

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$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

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The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

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As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

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As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

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