In a lab, .0092 moles of Cu were reacted with .0046 miles of S. What would happen to the mole Cu/mole S ratio if some of the copper was unreacted (increase, decrease, stay the same)?

Also, what will happen to the mole Cu/ Mole S ratio if unreacted sulfur remains in the crucible when the final weighing is made (increases, decreases, stays the same)?

To determine what would happen to the mole Cu/mole S ratio if some copper was unreacted, we need to first understand the balanced chemical equation for the reaction between copper (Cu) and sulfur (S):

Cu + S → CuS

From the equation, we can see that the stoichiometric ratio between Cu and S is 1:1. This means that for every 1 mole of copper reacting, 1 mole of sulfur is also consumed.

Given that .0092 moles of Cu and .0046 moles of S were reacted, we can calculate the mole Cu/mole S ratio:

Mole Cu/mole S = .0092 moles Cu / .0046 moles S = 2

Now, if some copper was unreacted, it would mean that less sulfur was consumed than expected based on the stoichiometric ratio. This would result in an increase in the mole Cu/mole S ratio. The ratio would become larger than 2 since the amount of copper remains the same while the amount of sulfur decreases.

Similarly, to determine what would happen to the mole Cu/mole S ratio if unreacted sulfur remains in the crucible when the final weighing is made, we need to consider that the unreacted sulfur will not contribute to the final product, which is copper sulfide (CuS).

If unreacted sulfur remains in the crucible, the mass of sulfur will be greater than initially measured. As a result, the mole Cu/mole S ratio will decrease. This is because the amount of sulfur in the denominator of the ratio will increase, while the amount of copper remains the same.

Is the question about the mole ratio in the product or the mixture??

If the product is CuS, the mole ratio in the product is 1molCu/1molS=1 molCu/molS
The mole ratio of the masses of the elements in any compound is definite, not dependent on the reactants ratio in making it.