Finding the moles required

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The discussion centers on calculating the moles of methyl iodide (CH3I) required to produce 0.1 moles of methanol (CH3OH) given a 50% yield in the reaction. The balanced chemical equation for the reaction is CH3I + H2O → CH3OH + HI. To achieve 0.1 moles of methanol at 50% yield, 0.2 moles of methyl iodide are necessary, as the theoretical yield must be doubled to account for the yield percentage.

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omzar
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ive been having a hard time with this question and any help would be appreciated.

If the percent yield in the reaction of methyl iodide (CH3I) with water to form methanol (CH3OH) and HI is 50%, how many moles of methyl iodide will be required to prepare 0.1 mole of methanol? (Assume that water is in excess.)
 
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Start with the reaction equation.
 

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