Question

How many grams of silver nitrate are necessary to react completely with 7.000 moles of copper?

Cu + 2 AgNO3 → Cu(NO3 )2 + 2 Ag
Select one:
a. 2379 g
b. 238 g
c. 14 moles
d. not enough info given

Answer 1

To react completely with 7.000 moles of Cu, you would need 14.000 moles of AgNO3, which is equivalent to 2379 grams.

Step-by-step explanation:

According to the balanced chemical equation:

Cu + 2 AgNO3 → Cu(NO3 )2 + 2 Ag

We see that 2 moles of AgNO3 react with 1 mole of Cu. Therefore, to react completely with 7.000 moles of Cu, we need:

2 moles AgNO3 / 1 mole Cu * 7.000 moles Cu = 14.000 moles AgNO3

Finally, we can convert moles of AgNO3 to grams using the molar mass of AgNO3:

14.000 moles AgNO3 * 169.87 g/mol = 2379 g

Answer 2

To completely react with 7.000 moles of copper, 595 grams of silver nitrate are necessary.

Step-by-step explanation:

In order to determine the amount of silver nitrate necessary to react completely with 7.000 moles of copper, we can use the balanced chemical equation:

Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2 Ag

From the balanced equation, we can see that for every 2 moles of AgNO₃, 1 mole of copper reacts. Therefore, if we have 7.000 moles of copper, we will need half of that, or 3.500 moles of silver nitrate.

Since the molar mass of AgNO₃ is approximately 169.87 g/mol, the mass of silver nitrate needed would be: 3.500 mol * 169.87 g/mol = 594.545 g. Rounding to the nearest gram, the answer is 595 g.