if 0.600 moles of copper(II) sulfide is treated with 1.40 moles of nitric acid, how many moles of copper(II) nitrate could be produced?

CuS + 2HNO3 = Cu(NO3)2 + H2S

That should help, right?

To determine the number of moles of copper(II) nitrate that could be produced, we need to find the limiting reactant. The limiting reactant is the reactant that will be completely consumed and prevent any further reaction.

First, let's write the balanced equation for the reaction between copper(II) sulfide and nitric acid:

CuS + 2HNO3 → Cu(NO3)2 + H2S

From the equation, we can see that 1 mole of copper(II) sulfide reacts with 2 moles of nitric acid to produce 1 mole of copper(II) nitrate.

The molar ratio between copper(II) sulfide and copper(II) nitrate is 1:1. Therefore, if 0.600 moles of copper(II) sulfide are used, then 0.600 moles of copper(II) nitrate could be produced.

In this case, the 1.40 moles of nitric acid is in excess since the number of moles of copper(II) sulfide is less than the number of moles of nitric acid required to react completely.

To determine the number of moles of copper(II) nitrate that could be produced, we first need to identify the balanced chemical equation for the reaction between copper(II) sulfide (CuS) and nitric acid (HNO3).

The balanced chemical equation for this reaction is:

CuS + 2HNO3 -> Cu(NO3)2 + H2S

According to the balanced equation, 1 mole of copper(II) sulfide reacts with 2 moles of nitric acid to produce 1 mole of copper(II) nitrate and 1 mole of hydrogen sulfide.

Given that we have 0.600 moles of copper(II) sulfide and 1.40 moles of nitric acid, we need to determine which reagent is limiting, meaning which one will be completely consumed and limit the amount of the product that can be produced.

To do this, we compare the moles of both reagents to their stoichiometric coefficients (the numbers in front of the compounds in the balanced equation).

For copper(II) sulfide, the stoichiometric coefficient is 1, and for nitric acid, it is 2.

Moles of copper(II) sulfide: 0.600 moles
Moles of nitric acid: 1.40 moles

To calculate the moles of copper(II) nitrate produced, we need to determine the limiting reagent. This can be done by comparing the moles of each reagent to their respective stoichiometric coefficients.

For copper(II) sulfide:
0.600 moles / 1 = 0.600 moles

For nitric acid:
1.40 moles / 2 = 0.700 moles

From the calculations above, we can see that the number of moles of copper(II) sulfide (0.600 moles) is less than the number of moles of nitric acid (0.700 moles). This means that copper(II) sulfide is the limiting reagent.

Since copper(II) nitrate has a 1:1 molar ratio with copper(II) sulfide, the number of moles of copper(II) nitrate that can be produced is also 0.600 moles.

Therefore, 0.600 moles of copper(II) nitrate could be produced.