Calculation of Volume of Methane Evolved

Given:
Mass of methyl magnesium iodide = 16.6 g
Conditions: S.T.P. (standard temperature and pressure)

Step 1:
Write the balanced chemical equation for the reaction between methyl magnesium iodide and water.
CH3MgI + H2O → CH4 + Mg(OH)I

Step 2:
Calculate the number of moles of methyl magnesium iodide using its molar mass.
Molar mass of CH3MgI = 142.28 g/mol
Number of moles of CH3MgI = mass/molar mass = 16.6 g/142.28 g/mol = 0.1168 mol

Step 3:
Use the stoichiometric coefficients from the balanced equation to find the number of moles of methane produced.
From the balanced equation, 1 mole of CH3MgI produces 1 mole of CH4.
Number of moles of CH4 = 0.1168 mol

Step 4:
Calculate the volume of methane at S.T.P using the ideal gas law.
PV = nRT, where P = 1 atm, V = ?, n = 0.1168 mol, R = 0.0821 L·atm/K·mol, T = 273.15 K
V = nRT/P = (0.1168 mol)(0.0821 L·atm/K·mol)(273.15 K)/(1 atm) = 2.24 L

Answer:
The volume of methane evolved by treatment of 16.6 g of methyl magnesium iodide with water at S.T.P is 2.24 L.

Explanation:
The given mass of methyl magnesium iodide is used to calculate the number of moles of the compound. From the balanced chemical equation, it is known that one mole of methyl magnesium iodide reacts to form one mole of methane. Using the ideal gas law, the volume of methane produced at S.T.P is calculated. The final answer is 2.24 L.

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