Question

What is the limiting reactant and how much excess reactant is leftover of:

4H2O+7CO2---> C7H8+9O2
starting with
50gH2O
250gCO2

Answer 1

Explanation: In a chemical reaction, the limiting reactant is the reactant that is completely consumed before the other reactant(s) run out. The limiting reactant determines the maximum amount of product that can be formed.

To determine the limiting reactant, we need to calculate the number of moles of each reactant present.

The moles of H2O: 50g/18g/mol = 2.778 moles

The moles of CO2: 250g/44g/mol = 5.68 moles

Next, we can use the coefficients in the balanced chemical equation to find the number of moles of each reactant required to react with each other.

The number of moles of H2O required: 4 moles H2O / 4 moles H2O/1 mole C7H8 = 1 mole C7H8

The number of moles of CO2 required: 7 moles CO2 / 1 mole C7H8 = 7 moles CO2

From the above we can see that H2O is the limiting reactant since it is consumed before the CO2.

To calculate the excess reactant, we subtract the number of moles of the limiting reactant from the number of moles of the other reactant.

The excess moles of CO2 = 5.68 moles - 1 mole = 4.68 moles

The excess moles of H2O = 0 moles (since H2O is limiting reactant)

The mass of the excess CO2 = 4.68 moles x 44 g/mol = 205.12 g

So, CO2 is the excess reactant and 205.12 g of CO2 is leftover.

Answer 2

The limiting reactant is the reactant that is completely consumed before the other reactant(s) in a chemical reaction. To find the limiting reactant, we need to find how much product is formed from each reactant, and then compare which reactant forms the least amount of product.

The balanced equation for the reaction is 4H2O+7CO2---> C7H8+9O2

From the given data, we have 50g of H2O and 250g of CO2.

We can calculate the limiting reactant by using the stoichiometry of the equation.

The stoichiometry is the relationship between the amount of reactants and products in a balanced chemical equation.

From the balanced equation, we know that 4 moles of H2O and 7 moles of CO2 are needed to produce 1 mole of C7H8 and 9 moles of O2.

From the data we have, 50g of H2O is equivalent to 2.86 moles H2O.

And 250g of CO2 is equivalent to 5.36 moles of CO2.

Now, using the balanced equation, we can calculate the amount of product that forms from each reactant:

From 2.86 moles of H2O: 2.86 moles H2O * (1 mole C7H8 / 4 moles H2O) = 0.715 moles C7H8

From 5.36 moles of CO2: 5.36 moles CO2 * (1 mole C7H8 / 7 moles CO2) = 0.766 moles C7H8

As we can see, the CO2 is the limiting reactant because it forms less product than H2O.

The amount of excess reactant is the amount of the reactant that is not used up in the reaction, we can calculate this by subtracting the amount of product that forms from the limiting reactant from the initial amount of the reactant.

Excess reactant of H2O is 50g - (0.715 moles * 18.02g/mol) = 37.5g

Excess reactant of CO2 is 250g - (0.766 moles * 44.01g/mol) = 191g

So, the limiting reactant is CO2 and there is 191g of excess CO2 leftover.