If 1.62 mold of CS2 burns with 5.65 mol of O2, how many moles of the excess reactant will still be present when the reaction is over? the chemical equation is CS2+3O2=2SO2+CO2
CS2 + 3O2 = 2SO2 + CO2
mols CS2 = 1.62
mols O2 = 5.65
How much CS2 is needed to burn 5.65 mols O2?
That's 5.65 mols O2 x (1 mol CS2/3 mols O2) = 5.65 x 1/3 = 1.88. You don't have that much; therefore, CS2 must be the limiting reagent but let's check to make sure.
How much O2 is need to use all of the CS2?
1.62 mols CS2 x (3 mols O2/1 mol CS2) = 1.62 x 3 = 4.86 and we have enough O2.
Therefore, 1.62 mols CS2 will burn using 4.86 mols O2. How much O2 is left? That's 5.65 - 4.86 = ? mols.
To determine the moles of the excess reactant, we need to first calculate the moles of each reactant used in the reaction.
From the balanced chemical equation: CS2 + 3O2 = 2SO2 + CO2, we can see that the stoichiometric ratio between CS2 and O2 is 1:3.
Given that 1.62 mol of CS2 is burned, we can calculate the moles of O2 required for complete combustion as follows:
Moles of O2 used = (Moles of CS2) * (Moles of O2 per mole of CS2)
Moles of O2 used = 1.62 mol * 3 mol O2/1 mol CS2
Moles of O2 used = 4.86 mol O2
Therefore, 4.86 mol of O2 would be completely used in the reaction.
Given that 5.65 mol of O2 is provided, we can calculate the moles of excess O2 remaining by subtracting the moles of O2 used from the moles of O2 provided:
Moles of excess O2 = (Moles of O2 provided) - (Moles of O2 used)
Moles of excess O2 = 5.65 mol - 4.86 mol
Moles of excess O2 = 0.79 mol
Hence, when the reaction is over, there will be 0.79 moles of excess O2 remaining.
To determine the moles of the excess reactant present after the reaction is over, we first need to identify the limiting reactant. The limiting reactant is the one that gets completely consumed in the reaction, thus limiting the amount of product that can be formed.
Let's calculate the number of moles of SO2 that can be formed from the given amounts of CS2 and O2:
From the balanced chemical equation: CS2 + 3O2 -> 2SO2 + CO2
We can see that 1 mole of CS2 reacts with 3 moles of O2 to produce 2 moles of SO2. Therefore, the molar ratio between CS2 and SO2 is 1:2.
Given that we have 1.62 moles of CS2, we can calculate the moles of SO2 that can be formed by multiplying the number of moles of CS2 by the molar ratio:
Moles of SO2 = 1.62 mol CS2 × (2 mol SO2 / 1 mol CS2) = 3.24 mol SO2
Next, we compare this calculated value to the amount of O2 available. Given that we have 5.65 moles of O2, we can calculate the moles of SO2 that can be formed by multiplying the number of moles of O2 by the molar ratio:
Moles of SO2 = 5.65 mol O2 × (2 mol SO2 / 3 mol O2) = 3.77 mol SO2
Since we have calculated that only 3.24 moles of SO2 can be formed from the given amount of CS2, which is less than the 3.77 moles of SO2 that can be formed from the given amount of O2, the limiting reactant in this case is CS2. This means that all of the CS2 will be consumed in the reaction, and there will be an excess of O2 remaining.
To find the amount of excess O2, we can subtract the amount of O2 consumed from the total amount of O2 provided:
Excess O2 = Total moles of O2 - Moles of O2 consumed
Excess O2 = 5.65 mol O2 - (3 moles of O2 needed to react with 1.62 mol CS2 × (5.65 mol O2 / 3 mol O2)) = 5.65 mol O2 - 3.354 mol O2 = 2.296 mol O2
Therefore, there will be approximately 2.296 moles of excess O2 remaining when the reaction is over.