Question

Consider the reaction of phosphorus with water and iodine. 2P (s) 6H2O (l) 3I2 (s) 6HI (aq) 2H3PO3 (aq) Determine the limiting reactant in a mixture containing 92.8 g of P, 257 g of H2O, and 1.39×103 g of I2. Calculate the maximum mass (in grams) of hydroiodic acid, HI, that can be produced in the reaction. The limiting reactant is:

Answer 1

Answer:

P is the limiting reactant

There will be 1151.19 grams of HI produced

Explanation:

Step 1: The balanced equation

2P(s) + 6H2O(l) + 3I2(s) → 6HI(aq)+2H3PO3(aq)

Step 2: Data given

mass of P = 92.8 grams

mass of H2O = 257 grams

mass of I2 = 1.39*10³ grams

Molar mass of P=30.97 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of I2 = 253.81 g/mol

Molar mass of HI = 127.91 g/mol

Step 3: Calculate number of moles

Number of moles = mass / Molar mass

Number of moles of P = 92.9 grams / 30.97 g/mol = 3

Number of moles of H2O = 257 grams / 18.02 g/mol =14.26 moles

Number of moles of I2 =1390 grams / 253.81 g/mole = 5.477 moles

Step 4: Find the limiting reactant

For 2 moles P consumed, we need 6 moles of H2O consumed and 3 moles of I2 consumed to produce 6 moles of Hi and 2 moles H3PO3.

The limiting reactant is P. IT will completely react, so 3 moles. There will react 3* 3 = 9 moles of H2O. THere will remain 14.26 - 9 = 5.26 moles of H2O.

There will react 3*3/2 = 4.5 moles of I2. There will remain 5.477 - 4.5  = 0.977 moles I2

Step 5: Calculate mass of HI produced

For 2 moles P consumed, we need 6 moles of H2O consumed and 3 moles of I2 consumed to produce 6 moles of Hi.

For 3 moles of P consumed, we produce 3*3 = 9 moles of HI

mass HI = moles * Molar mass

mass HI = 9 moles * 127.91 g/mol = 1151.19 grams

There will be 1151.19 grams of HI produced