How many moles of iodomethane are consumed in | vedclass

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(B) The reaction of methylamine $(CH_3NH_2)$ with iodomethane $(CH_3I)$ proceeds via exhaustive alkylation until the quaternary ammonium salt is forme

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Three

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(B) The reaction of methylamine $(CH_3NH_2)$ with iodomethane $(CH_3I)$ proceeds via exhaustive alkylation until the quaternary ammonium salt is formed: $CH_3NH_2 + CH_3I \rightarrow (CH_3)_2NH + HI$ $(CH_3)_2NH + CH_3I \rightarrow (CH_3)_3N + HI$ $(CH_3)_3N + CH_3I \rightarrow (CH_3)_4N^{+}I^{-}$ Summing these steps,$3$ moles of $CH_3I$ are consumed to convert $1$ mole of $CH_3NH_2$ into tetramethylammonium iodide $((CH_3)_4N^{+}I^{-})$.

How many moles of iodomethane are consumed in the following conversion?
$CH_3NH_2 \xrightarrow[\Delta]{CH_3I} (CH_3)_4N^{+}I^{-}$

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How many moles of iodomethane are consumed in the following conversion?$CH_3NH_2 \xrightarrow[\Delta]{CH_3I} (CH_3)_4N^{+}I^{-}$