Calculation of Volume of Methane Evolved from Methyl Magnesium Iodide

Given:
Mass of methyl magnesium iodide = 16.6g

Steps to Calculate Volume of Methane:
1. Calculate the moles of methyl magnesium iodide using its molar mass.
2. Write the balanced chemical equation for the reaction of methyl magnesium iodide with water to form methane and magnesium hydroxide.
3. Use stoichiometry to determine the number of moles of methane produced.
4. Convert the number of moles of methane to volume using the ideal gas law.

Step 1: Calculate the Moles of Methyl Magnesium Iodide
The molar mass of methyl magnesium iodide (CH3MgI) = 141.32 g/mol.

Number of moles of CH3MgI = mass/molar mass
Number of moles of CH3MgI = 16.6g/141.32 g/mol
Number of moles of CH3MgI = 0.1175 mol

Step 2: Write the Balanced Chemical Equation
CH3MgI + H2O → CH4 + Mg(OH)I

Step 3: Use Stoichiometry to Determine the Number of Moles of Methane Produced
From the balanced chemical equation, 1 mole of CH3MgI reacts with 1 mole of H2O to produce 1 mole of CH4.

Therefore, the number of moles of CH4 produced = 0.1175 mol.

Step 4: Convert Moles of Methane to Volume
The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Assuming standard temperature and pressure (STP), which is 1 atm and 273 K, we can calculate the volume of methane produced.

V = nRT/P
V = (0.1175 mol)(0.0821 L atm/mol K)(273 K)/1 atm
V = 2.94 L

Therefore, the volume of methane evolved by treatment of 16.6g of methyl magnesium iodide with water is 2.94 L.

2.24l