What are the answers to the three questions? And how do you calculate and get those answers?

Question

0.98 g of silver (I) nitrate was dissolved in water and a 0.56 g piece of copper wire was placed in the solution. The color due to silver (I) nitrate soon faded and a precipitate of solid silver was observed on the wire. After the color of the solution had disappeared completely, the leftover copper wire was removed from the solution and weighed. The mass of the leftover copper wire was 0.43 g. 

The balanced equation is 2AgNO3 + Cu --> Cu(NO3)2 + 2Ag

The limiting reactant is 2AgNO3. 

1. How much of the excess reactant is left over after the reaction goes to completion?

1. Using the limiting reactant, how much of each product actually produced.

1. If the reaction was only 97.65% efficient, how much of each product is actually produced?

Answer 1

Hello, Nicole,

How much of the excess reactant is left over after the reaction goes to completion?

Since the limiting reagent is the AgNO₃, the mass of the "leftover" copper wire (0.43 grams) is the excess reactant.

Using the limiting reactant, how much of each product actually produced.

The balanced equation tells us the ratio of each product relative to the silver nitrate.

2AgNO₃ + Cu --> Cu(NO₃)₂ + 2Ag

1. For Cu(NO₃)₂ the ratio of product to reactant is the ratio of their coefficients in the balanced equation. We get 1 mole Cu(NO₃)₂ per 2 moles of AgNO₃, a molar ratio of 1/2.
2. For Ag, we get 2 moles Ag per 2 moles AgNO₃, a molar ratio of 1

These ratios can then be multiplied by the actual moles of AgNO₃ to determine the moles of products we would expect. To get there, convert 0.98 gram of sliver nitrate into moles by dividing by it's molar mass of 169.9 grams/mole. That equals 0.0058 moles. Now multiply this mole quantity by each of the molar ratios to find how much of each product we'd expect. Note that the mass of copper (0.56 g) makes 0.088 moles of copper, more than 10 times the moles we have of the silver nitrate. The equation says we only need twice as much silver nitrate as copper. Thus, we have much more copper than is needed and so the reaction will stop once the silver nitrate is consumed (the limiting reagent)

For Cu(NO₃)₂ the amount produced is (0.0058 moles AgNO₃)*(1 mole Cu(NO₃)₂/2 moles AgNO₃) = 0.0029 moles Cu(NO₃)₂ Convert this to grams by multiplying by the molar mass of Cu(NO₃)₂, which is 187.6 grams/mole. This gives 1.65 grams Cu(NO₃)₂.

For Ag, we do the same calculation as above, but now with a molar ratio of 1, we know we should expect that same moles of Ag as we consumed of the Cu(NO₃)_2, 0.0058 moles. At a molar mass of 107.9 g/mole, this says we would hope to find 0.622 grams of silver produced. [A meager reward for all this work].

If the reaction was only 97.65% efficient, how much of each product is actually produced?

Take the amounts calculated above and multiply by 0.9765 to find the amount actually produced.

Ag: 0.622*(0.9765) = 0.607 grams

Cu(NO₃)₂ : 1.65*(0.9765) = 1.61 grams

I hope this helps,

Bob