5C(s)+2SO2(g)->CS2(I)+4CO(g)

A-) if you have 6.78 g of C and 7.54 g of SO2, how many grams of CO can be created?
B-) what is the limiting reagent from A?
C-) how much of the excess reagent is left after reaction?
D-) if only 5.89 of CO was actually created, what is the percent yield?

To solve the given problems, we need to use stoichiometry and the concept of limiting reagents. Let's go step by step:

A-) To find the grams of CO produced, we first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed, thus dictating the maximum amount of product that can be formed.

1. Calculate the number of moles for each reactant:
- Moles of C = mass/Molar mass = 6.78 g/12.01 g/mol of C
- Moles of SO2 = mass/Molar mass = 7.54 g/64.07 g/mol of SO2

2. Use the balanced chemical equation to determine the stoichiometric ratio between C and CO:
- According to the balanced equation, one mole of C produces four moles of CO.

3. Compare the moles of reactants obtained in step 1:
- To find the limiting reagent, we need to compare the mole ratio between C and SO2. Divide the moles of C by the coefficient in front of C in the balanced equation, and do the same for SO2.

Based on the calculation, we can see that the ratio of moles of C to SO2 is 5.95/1.18 = 5.04. Since the mole ratio is greater than 4 (the stoichiometric ratio of C to CO), SO2 is the limiting reagent.

4. Calculate the moles of CO produced from the limiting reagent:
- From the balanced equation, we know that four moles of CO are produced from one mole of SO2.

Now we can calculate the moles of CO produced:
Moles of CO = moles of SO2 x (4 moles of CO / 2 moles of SO2)

Finally, we can find the grams of CO produced by multiplying the moles of CO by its molar mass.

B-) From the previous calculations, we determined that SO2 is the limiting reagent.

C-) To find the amount of excess reagent remaining after the reaction, we need to determine the moles of an excess reagent and then convert it back to grams.

1. Calculate the moles of the excess reagent (which is C in this case) by multiplying the moles of C by its molar mass.

2. Subtract the moles of the excess reagent (C) consumed from its original moles to find the moles of the excess reagent remaining.

3. Convert the moles of the excess reagent remaining to grams by multiplying it by the molar mass of C.

D-) To calculate the percent yield, we need to compare the actual yield (5.89 g of CO) to the theoretical yield (which we will calculate using the limiting reagent).

1. Calculate the theoretical yield of CO using the limiting reagent (SO2).
- Convert the moles of the limiting reagent (SO2) to grams of CO using the molar mass of CO.

2. Calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100%

Now you have the step-by-step explanation for solving questions A, B, C, and D related to the given chemical equation.

What is your hang up on this question?