Respuesta :
Answer : The mass of silver chloride formed are, 8.44 grams.
Explanation : Given,
Mass of [tex]AgNO_3[/tex] = 10.0 g
Mass of [tex]BaCl_2[/tex] = 15.0 g
Molar mass of [tex]AgNO_3[/tex] = 169.87 g/mole
Molar mass of [tex]BaCl_2[/tex] = 208.23 g/mole
Molar mass of [tex]AgCl[/tex] = 143.32 g/mole
First we have to calculate the moles of [tex]AgNO_3[/tex] and [tex]BaCl_2[/tex].
[tex]\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{10g}{169.87g/mole}=0.0589moles[/tex]
[tex]\text{Moles of }BaCl_2=\frac{\text{Mass of }BaCl_2}{\text{Molar mass of }BaCl_2}=\frac{15g}{208.23g/mole}=0.072moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2AgNO_3+BaCl_2\rightarrow 2AgCl+Ba(NO_3)2[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]AgNO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]
So, 0.0589 moles of [tex]AgNO_3[/tex] react with [tex]\frac{0.0859}{2}=0.0295[/tex] moles of [tex]BaCl_2[/tex]
From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]AgCl[/tex].
As, 2 moles of [tex]AgNO_3[/tex] react to give 2 moles of [tex]AgCl[/tex]
So, 0.059 moles of [tex]AgNO_3[/tex] react to give 0.059 moles of [tex]AgCl[/tex]
Now we have to calculate the mass of [tex]AgCl[/tex].
[tex]\text{Mass of }AgCl=\text{Moles of }AgCl\times \text{Molar mass of }AgCl[/tex]
[tex]\text{Mass of }AgCl=(0.0589mole)\times (143.32g/mole)=8.44g[/tex]
Therefore, the mass of silver chloride formed are, 8.44 grams.
The mass of silver chloride (AgCl) formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride is 8.44 g. The correct option is the second option 8.44 g
First, we will write a balanced chemical equation for the reaction between silver nitrate and barium chloride
2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
This means, 2 moles of AgNO₃ will react with 1 mole of BaCl₂ to produce 2 moles of AgCl and 1 mole of Ba(NO₃)₂
From the question,
10.0 g of silver nitrate reacts with 15.0 g of barium chloride
First, we will determine the number of moles present in each reactant
- For silver nitrate (AgNO₃)
Mass = 10.0 g
Molar mass = 169.87 g/mol
Using the formula
[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]
Then,
[tex]Number \ of\ moles\ of\ silver\ nitrate = \frac{10.0}{169.87}[/tex]
Number of moles of AgNO₃ = 0.05887 moles
- For Barium chloride (BaClâ‚‚)
Mass = 15.0 g
Molar mass of BaClâ‚‚ = 208.23 g/mol
∴ [tex]Number \ of\ moles\ of\ barium \ chloride= \frac{15.0}{208.23}[/tex]
Number of moles of BaClâ‚‚ = 0.072036 moles
From the balanced chemical equation
2 moles of AgNO₃ will react with 1 mole of BaCl₂
∴ 0.05887 moles AgNO₃ will react with [tex]\frac{0.05887}{2}[/tex] mole of BaCl₂
[tex]\frac{0.05887}{2} = 0.029435[/tex]
This means only 0.029435 moles of BaClâ‚‚ will react
(NOTE: AgNO₃ is the limiting reagent while BaCl₂ is the excess reagent)
Now,
Since 2 moles of AgNO₃ will react with 1 mole of BaCl₂ to produce 2 moles of AgCl
That means,
0.05887 moles AgNO₃ will react with 0.029435 moles of BaCl₂ to produce 0.05887 moles of AgCl
∴ The number of moles of silver chloride (AgCl) produced is 0.05887 moles
Now, to determine the mass (in grams) of silver chloride that are formed
From the formula,
Mass = Number of moles × Molar mass
Molar mass of AgCl = 143.32 g/mol
∴ Mass of silver chloride, AgCl, formed = 0.05887 moles × 143.32 g/mol
Mass of silver chloride formed = 8.4372484 g
Mass of silver chloride formed ≅ 8.44 g
Hence, the mass of silver chloride (AgCl) formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride is 8.44 g. The correct option is the second option 8.44 g
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