2C8H18(g)+25O2(g)-->16CO2(g)+18H2O(g)

0.290mol of octane is allowed to react with 0.600mol of oxygen.The limiting reactant is Oxygen.
How many moles of water are produced in this reaction?

After the reaction, how much octane is left?

0.600 mole oxygen x (18 moles H2O/25 moles O2) = ?? moles H2O produced.

Use the same logic to convert 0.600 moles oxygen to moles octane, convert moles octane to grams octane, subtract from the grams you started with (of course convert 0.290 mols octane to grams) to determine grams remaining unreacted.

To find out how many moles of water are produced in the reaction, we can use the stoichiometric ratio from the balanced equation. According to the equation:

2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)

We can see that for every 2 moles of octane (C8H18) reacted, 18 moles of water (H2O) are produced. Therefore, we can set up a ratio:

2 moles of C8H18 reacts to produce 18 moles of H2O

From the information given, we know that 0.290 moles of octane (C8H18) is reacted. So, we can calculate the moles of water produced as follows:

0.290 moles C8H18 x (18 moles H2O/2 moles C8H18) = 2.61 moles H2O

Therefore, 2.61 moles of water are produced in this reaction.

To determine how much octane is left after the reaction, we need to compare the moles of oxygen and octane used. From the given information, we know that 0.600 moles of oxygen (O2) is used, and we have already determined that oxygen is the limiting reactant.

The stoichiometric ratio tells us that 25 moles of oxygen (O2) reacts with 2 moles of octane (C8H18). So, we can calculate the moles of octane reacted as follows:

0.600 moles O2 x (2 moles C8H18/25 moles O2) = 0.048 moles C8H18

Therefore, 0.048 moles of octane are reacted. To find out how much octane is left, we can subtract the moles reacted from the initial moles of octane:

0.290 moles C8H18 - 0.048 moles C8H18 = 0.242 moles C8H18

Therefore, after the reaction, 0.242 moles of octane are left.

To find out how many moles of water are produced in the reaction, we need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed first and determines the maximum amount of product that can be formed.

In this reaction, we are given that 0.290 mol of octane (C8H18) and 0.600 mol of oxygen (O2) are present. We need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.

From the balanced equation: 2C8H18(g) + 25O2(g) -> 16CO2(g) + 18H2O(g)

The stoichiometric coefficient for octane is 2, and for oxygen, it is 25. We can calculate the number of moles of water that can be produced from the amount of oxygen present. Since the coefficient of oxygen is higher, we will consider it as the limiting reactant.

Using the ratio of the coefficients, we can set up a proportion:

25 mol of oxygen (O2) -> 18 mol of water (H2O)
0.600 mol of oxygen (O2) -> x moles of water (H2O)

x = (0.600 mol O2) * (18 mol H2O / 25 mol O2)
x = 0.432 mol H2O

Therefore, 0.432 mol of water are produced in this reaction.

To find out how much octane is left after the reaction, we need to determine the number of moles of octane that reacted. Since we know the limiting reactant is oxygen, we need to calculate the moles of octane that would react completely with 0.600 mol of oxygen.

Using the stoichiometric coefficients from the balanced equation:

25 mol of oxygen (O2) -> 2 mol of octane (C8H18)
0.600 mol of oxygen (O2) -> x moles of octane (C8H18)

x = (0.600 mol O2) * (2 mol C8H18 / 25 mol O2)
x = 0.048 mol C8H18

Therefore, 0.048 mol of octane is consumed, which means the remaining amount of octane is:

0.290 mol - 0.048 mol = 0.242 mol of octane

So, after the reaction, 0.242 mol of octane is left.

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