It takes 45.60 mL of a 0.225 M hydrochloric acid solution to react completely with 25.00 mL of calcium hydroxide in this reaction below, what is the molar concentration of the calcium hydroxide solution?


2HCl(aq) + Ca(OH)2(aq)!CaCl2(aq) + 2H2O(l)

Answers

Answer 1

Given :

It takes 45.60 mL of a 0.225 M hydrochloric acid solution to react completely with 25.00 mL of calcium hydroxide in the given reaction.

2HCl(aq) + Ca(OH)2(aq) --> CaCl2(aq) + 2H2O(l)

To Find :

The molar concentration of the calcium hydroxide solution.

Solution :

From given equation 2 mole of hydrochloric acid react with 1 mole of calcium hydroxide.

So,

[tex]2M_1V_1 = M_2V_2\\\\2\times 0.225 \times 45.60 = M_2 \times 25 \\\\M_2 = \dfrac{20.52}{25}\\\\M_2 = 0.8208 \ M[/tex]

Therefore, molar concentration of the calcium hydroxide solution is 0.8208 M.

Answer 2

Answer:

Molar concentration of the calcium hydroxide solution = 0.2052 M

Explanation:

Volume of HCl = 45.60 mL  = 0.0456 L

Molar concentration of HCl = 0.225 M

Volume of Ca(OH)2 = 25.00 mL  = 0.025 L

No. of moles of HCl = (0.225 M)*(0.0456 L)   = 0.01026 mol

As we know

2 moles of HCl require 1 mole of Ca(OH)2, hence, .01026 moles of HCl will need 0.5*(0.01026) = 0.00513 moles of Ca(OH)2

Molar concentration of Ca(OH)2 = [(0.00513 mol)/(0.025 L)]   = 0.2052 M


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Answers

Answer:

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Answers

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Explanation: I added a little more info about them please give me brainliest


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Answers

Answer:  

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2) [tex]Q_{rxn}=4435.04J[/tex]  

3) [tex]\Delta _rH=15.8kJ/mol[/tex]

Explanation:  

Hello there!  

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.  

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

[tex]Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J[/tex]    

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

[tex]\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol[/tex]

Best regards!  

Best regards!

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