Answer:
[tex]m_{CH_3OH}=4.70x10^{-20} gCH_3OH[/tex]
Explanation:
Hello.
In this case, since the balanced chemical reaction is:[tex]2H_2(g) + CO(g) \rightarrow CH_3OH(l)[/tex]
By starting with 885 molecules of carbon monoxide, we can compute the grams of methanol (molar mass: 32 g/mol) by using the Avogadro's number, the 1:1 mole ratio of CO to methanol and methanol's molar mass as shown below:
[tex]m_{CH_3OH}=885 molec\ CO*\frac{1molCO}{6.022x10^{23}molec\ CO} *\frac{1molCH_3OH}{1molCO} *\frac{32gCH_3OH}{1molCH_3OH} \\\\m_{CH_3OH}=4.70x10^{-20} gCH_3OH[/tex]
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