the reactant that is completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed. To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometry of the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction between P, H2O, and I2 to form HI:
P + 4H2O + 2I2 -> 4HI + H3PO4
From the balanced equation, we can see that the stoichiometric ratio between P and HI is 1:4. This means that for every 1 mole of P, we can produce 4 moles of HI.
To determine the limiting reactant, we need to convert the masses of P, H2O, and I2 to moles. We can use the molar masses of each substance to do this.
The molar mass of P is 30.97 g/mol.
The molar mass of H2O is 18.02 g/mol.
The molar mass of I2 is 253.80 g/mol.
To convert the masses to moles, we divide each mass by its respective molar mass:
Number of moles of P = 197 g / 30.97 g/mol
Number of moles of H2O = 402 g / 18.02 g/mol
Number of moles of I2 = 2.07×103 g / 253.80 g/mol
Now that we have the number of moles of each reactant, we can compare them to the stoichiometry of the balanced equation.
Let's assume that the number of moles of P is x. Then, the number of moles of HI produced would be 4x.
From the stoichiometry, we know that the number of moles of H2O is 4 times the number of moles of P, and the number of moles of I2 is 2 times the number of moles of P.
So, the number of moles of H2O is 4x, and the number of moles of I2 is 2x.
To determine the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometry of the balanced equation.
The limiting reactant is the reactant that produces the smallest amount of product. In this case, we need to find the reactant that produces the smallest number of moles of HI.
To do this, we can compare the number of moles of HI produced by each reactant.
Number of moles of HI produced by P = 4x
Number of moles of HI produced by H2O = 4(4x) = 16x
Number of moles of HI produced by I2 = 2(4x) = 8x
From the above calculations, we can see that the number of moles of HI produced by P is 4x, the number of moles of HI produced by H2O is 16x, and the number of moles of HI produced by I2 is 8x.
Since we want to find the limiting reactant, we need to find the reactant that produces the smallest number of moles of HI. Therefore, we need to find the value of x that minimizes the number of moles of HI produced.
To do this, we can set up the following inequality:
4x ≤ 8x ≤ 16x
Simplifying the inequality, we get:
4x ≤ 8x ≤ 16x
Dividing all terms by x, we get:
4 ≤ 8 ≤ 16
From this inequality, we can see that the smallest value of x that satisfies the inequality is x = 4.
Therefore, the limiting reactant is P, and the number of moles of HI produced by P is 4x = 4(4) = 16 moles.
To calculate the maximum mass of HI that can be produced, we can use the molar mass of HI, which is 127.91 g/mol.
The maximum mass of HI that can be produced is given by:
Mass of HI = Number of moles of HI × Molar mass of HI
Mass of HI = 16 moles × 127.91 g/mol
Now, we can calculate the maximum mass of HI.