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kolbaska11 [484]
3 years ago
15

The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of AgNO_{3} in 50.0 mL of 0.100 M of AgNO_{3}

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= \frac{0.100}{1000}\times 50.0 moles

= 0.00500 moles

According to balanced equation, 1 mol of AgNO_{3} reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of AgNO_{3} react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = (100.0\times 1.00) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= \frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}

= \frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}

= -67716 J/mol

[m = mass, c = specific heat capacity, \Delta T = change in temperature and negative sign is included as it is an exothermic reaction]

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Answer:

39.6 g

Explanation:

The equation of the reaction is;

2Mg(s) + O2(g) --------> 2MgO(s)

To obtain the limiting reactant;

Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles

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1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO

Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles

If 1 mole of O2 yields 2 moles of MgO

0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO

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Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g

Percent yield = 90%

Percent yield = actual yield/theoretical yield * 100

Actual yield = Percent yield * theoretical yield/100

Actual yield = 90 * 44/100

Actual yield = 39.6 g

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