Respuesta :
Explanation:
In order to calculate the number of moles of a compound, we will find it out as follows.
No. of moles of chromium(III) chloride = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{158.36}[/tex]
= 0.63 moles
No. of moles of [tex]NH_{3}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{17}[/tex]
= 5.88 moles
No. of moles of [tex]MgCl_{2}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{95}[/tex]
= 1.05 moles
No. of moles of [tex]H_{3}PO_{4}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{98}[/tex]
= 1.02 moles
No. of moles of KCl = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{74.5}[/tex]
= 1.34 moles
No. of moles of [tex]N_{2}O_{3}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{76}[/tex]
= 1.31 moles
No. of moles of [tex]CH_{4}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{16}[/tex]
= 6.25 moles
No. of moles of [tex]C_{3}H_{8}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{44}[/tex]
= 2.27 moles
No. of moles of [tex]LiClO_{3}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{100}{90.39}[/tex]
= 1.11 moles
Therefore, we can conclude that out of the given options [tex]CH_{4}[/tex] contains the largest number of moles of compound.
