Question

When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The balanced equation is:
2AgNO3(aq) + Na2SO4(aq) → Ag2SO4(s) + 2NaNO3(aq)

How many grams of silver sulfate are expected when a solution containing 0.20 mol AgNO3 is mixed with a solution containing 0.15 mol Na2SO4?

Answer 1

Answer:

31.2g of Ag2SO4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AgNO3(aq) + Na2SO4(aq) → Ag2SO4(s) + 2NaNO3(aq)

Step 2:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

2 moles of AgNO3 reacted with 1 mole of Na2SO4.

Therefore, 0.2 mole of AgNO3 will react with = (0.2 x 1)/2 = 0.1 mole of Na2SO4.

From the above calculation, we can see that there are leftover for Na2SO4 as only 0.1 mole reacted out of 0.15 mole that was given. Therefore, AgNO3 is the limiting reactant.

Step 3:

Determination of the number of mole of Ag2SO4 produced from the reaction.

The limiting reactant is used to obtain the maximum product of the reaction.

2AgNO3(aq) + Na2SO4(aq) → Ag2SO4(s) + 2NaNO3(aq)

From the balanced equation above,

2 moles of AgNO3 produced 1 mole of Ag2SO4.

Therefore, 0.2 mole of AgNO3 will produce = (0.2 x1) /2 = 0.1 mole of Ag2SO4.

Step 4:

Conversion of 0.1 mole of Ag2SO4 to gram. This is illustrated below:

Molar Mass of Ag2SO4 = (2x108) + 32 + (16x4) = 216 + 32 + 64 = 312g/mol

Number of mole of Ag2SO4 = 0.1 mole

Mass of Ag2SO4 =?

Mass = number of mole x molar Mass

Mass of Ag2SO4 = 0.1 x 312

Mass of Ag2SO4 = 31.2g

Therefore, 31.2g of Ag2SO4 is produced from the reaction.