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jarptica [38.1K]
4 years ago
12

A 1.00 x 102mL sample of 0.200 M aqueous hydrochloric acid is added to 1.00 x 102mL of 0.200 M aqueous ammonia in a constant-pre

ssure calorimeter of negligible heat capacity. The following reaction occurs when the two solutionsare mixedHCl(aq)+ NH3(aq)--->NH4Cl(aq)The temperature increase is 2.34°C. Calculate heat change of the reaction per mole of HCl reacted.Assume that the densities and specific heats of the solutions are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively)
Chemistry
1 answer:
insens350 [35]4 years ago
5 0

Answer:

ΔH = -98 kJ/mol

Explanation:

<u>To calculate the heat change of the reaction:</u>

HCl(aq) + NH₃(aq)  →  NH₄Cl(aq)      

 0.2M        0.2M        ΔT=2.34°C  

1x10²mL   1x10²mL          

<u>We need to use the next equation:</u>

q = mc \Delta T (1)

<em>where </em><em>q: the amount of heat energy lost or gained</em><em>, </em><em>m: the mass of the substance, c: the specific heat capacity of the substance and ΔT: the change in temperature of the substance      </em>              

<u>Assuming that the densities of the solutions are the same as for water, we can determine the mass of the solution:</u>

d = \frac {m}{V}                    

<em>where </em><em>d: density, m: mass and V: volume of solution</em><em> </em><em>= 100 + 100 = 200mL</em><em> </em>

m = d \cdot V = 1 \frac {g}{mL} \cdot 200mL = 200g      

<u>Now, using the calculated mass in equation</u><u> (1)</u><u>, and assuming that the specific heats of the solutions are the same as for water, we can find heat change of the reaction</u>:      

q = 200g \cdot 4.184 \frac {J}{g \cdot ^{\circ}C} \cdot 2.34^{\circ}C      

q = 1.96 \cdot 10^{3}J          

This heat is negative because is the heat lost by the reacting HCl and NH₃ and gained by the water, so:

q = - 1.96 \cdot 10^{3}J    

<u>To calculate the heat change of the reaction per mole of HCl, we need to divide the heat change by the number of moles, which is called the enthalpy of reaction:</u>

\Delta H = \frac {q}{moles}

\Delta H = \frac {q}{moles}  

\Delta H = \frac {-1.96 \cdot 10^{3} J}{0.2 \frac{mol}{L} \cdot 0.1L}

\Delta H = -98 \cdot 10^{3} \frac{J}{mol} = -98 \frac {kJ}{mol}

So, the heat change of the reaction per mole of HCl reacted, often called enthalpy of reaction, is ΔH = -98 kJ/mol.    

Have a nice day!

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Following are the steps to find theoretical yield .

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Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

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b) To find mole ratio of AlCl₃ : Al

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Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

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Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

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Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

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Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

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Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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