Answer:
16.0 mol H.
Explanation:
Hello!
In this case, since the Avogadro's number help us to compute the moles of methanol in the given molecules:
[tex]n_{CH_3OH}=2.41x10^{24}molecCH_3OH*\frac{1molCH_3OH}{6.022x10^{23}molecCH_3OH}=4.00molCH_3OH[/tex]
Thus, since 1 mole of methanol contains 4 moles of hydrogen atoms, we obtain:
[tex]n_H=4.00molCH_3OH*\frac{4molH}{1molCH_3OH}\\\\n_H=16.0molH[/tex]
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