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mezya [45]
3 years ago
9

How many grams of silver chloride can be prepared by the reaction of 105.0 ml of 0.22 m silver nitrate with 105.0 ml of 0.13 m c

alcium chloride?
Chemistry
1 answer:
Ksenya-84 [330]3 years ago
6 0
The balanced equation for the above reaction is as follows;
2AgNO₃ + CaCl₂ ---> 2AgCl + Ca(NO₃)₂
stoichiometry of AgNO₃ to CaCl₂ is 2:1
we first need to find the limiting reactant 
number of moles reacted = molarity x volume
number of AgNO₃ moles = 0.22 mol/L x 0.1050 L = 0.023 mol
number of CaCl₂ moles = 0.13 mol/L x 0.1050 L = 0.014 mol
according to molar ratio of 2:1
if we assume AgNO₃ to be the limiting reactant 
if 2 mol of AgNO₃ react with 1 mol of CaCl₂
then 0.023 mol of AgNO₃ reacts with - 0.023/2 = 0.012 mol of CaCl₂
0.012 mol of CaCl₂ is required but 0.014 mol of CaCl₂ is required
therefore CaCl₂ is in excess and AgNO₃ is therefore the limiting reactant 

the amount of products formed depends on the amount of limiting reactant present 
stoichiometry of AgNO₃ to AgCl is 2:2
the number of moles of AgCl formed = number of AgNO₃ moles reacted 
therefore number of AgCl moles formed = 0.023 mol
mass of AgCl formed = 0.023 mol x 143.3 g/mol = 3.3 g
mass of AgCl formed = 3.3 g
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4 0
4 years ago
Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t
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Answer:

\Delta G^0 _{rxn} = 207.6\ kJ/mol

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

\Delta G_f^0  \ \ \ N_2O_{(g)} = 103 .8  \ kJ/mol

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\Delta G_f^0  \ \ \ O_2{(g)} =0 \ kJ/mol

\Delta G^0 _{rxn} = 2 \times \Delta G_f^0  \ N_2O_{(g)} - ( 2 \times  \Delta G_f^0  \ N_2{(g)} +   \Delta G_f^0  \ O_{2(g)})

\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times  0 +   0)

\Delta G^0 _{rxn} = 207.6\ kJ/mol

The equilibrium constant determined from the partial pressure denoted as K_p can be expressed as :

K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}

K_p = \dfrac{1}{ (22.20)}

K_p = 0.045

\Delta G = \Delta G^0 _{rxn} + RT \ lnK

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R = gas constant = 8.314 × 10⁻³ kJ

\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15  \ ln(0.045)

\Delta G =207.6 + 2.4788191 \times \ ln(0.045)

\Delta G =207.6+ (-7.687048037)

\Delta G = 199.912952  kJ

ΔG ≅ 199.91 kJ

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4 years ago
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