Answer:
Option B. 2.62g.
Explanation:
We'll begin by calculating the masses of AgNO3 and NaCl that reacted and the mass of AgCl produced from the balanced equation. This is illustrated below:
AgNO3(aq) + NaCl(aq) —> AgCl(s) + NaNO3(aq)
Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol
Mass of AgNO3 from the balanced equation = 1 x 170 = 170g
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass of NaCl from the balanced equation = 1 x 58.5 = 58.5g
Molar mass of AgCl = 108 + 35.5 = 143.5g/mol
Mass of AgCl from the balanced equation = 1 x 143.5 = 143.5g
Summary:
From the balanced equation above,
170g of AgNO3 reacted with 58.5g of NaCl to 143.5g of AgCl.
Next, we shall determine the limiting reactant. This can be obtain as follow:
From the balanced equation above,
170g of AgNO3 reacted with 58.5g of NaCl.
Therefore, 3.10g of AgNO3 will react with = (3.10 x 58.5)/170 = 1.07 g of NaCl
From the calculations made above,
only 1.07g out of 3.10g of NaCl given reacted completely with 3.10g of AgNO3.
Therefore, AgNO3 is the limiting reactant and NaCl is the excess reactant.
Now, we can calculate the mass of AgCl produced from the reaction of 3.10g of AgNO3 and 3.10g of NaCl as illustrated below:
In this case, the limiting reactant will be use as it will produce the maximum yield of AgCl since all of it is consumed in the reaction.
The limiting reactant is AgNO3 and the mass of AgCl produced can be obtained as follow:
From the balanced equation above,
170g of AgNO3 reacted to produce 143.5g of AgCl.
Therefore, 3.10g of AgNO3 will react to produce = (3.10 x 143.5)/170 = 2.62g of AgCl.
Therefore, 2.62g of AgCl were produced from the reaction.
