Consider the combination reaction of samarium metal and oxygen gas. if you start with 33.7 moles of samarium metal, how many moles of oxygen gas would be required to react completely with all of the samarium metal? for this reaction, samarium has a +3 oxidation state within the samarium/oxygen compound.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of  samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

of oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

i do not get it   at all