Respuesta :
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
