According to the reaction shown, how many moles of oxygen gas are necessary to produce 24.0 g of Li2O?

4Li(s)+O2(s)-2Li2O(s)

Any help would be appreciated. If someone could walk me through the entire thing as if I know nothing about chemistry, that would help because I'm completely lost in this class right now.

Question

According to the reaction shown, how many moles of oxygen gas are necessary to produce 24.0 g of Li2O?

4Li(s)+O2(s)->2Li2O(s)

Any help would be appreciated. If someone could walk me through the entire thing as if I know nothing about chemistry, that would help because I'm completely lost in this class right now.

anonymous · Answer

Alright. First step, we need the amount of $Li_2O$. To calculate the amount of $Li_2O$, we use the formula $\large n_{Li_2O}=\frac{mass}{molar~mass}$.

Next the equation is      $4Li~~~~~~~+~~~~~~~O_2~~~~~~~\rightarrow~~~~~~~2Li_2O$
                                                            1 moles             2 moles
                                                             ? moles             $n$ moles 

Therefore, $\Large n_{O_2}=\frac{n_{Li_2O} \times 1}{2}=\frac{n_{Li_2O}}{2}$. Plug $n$ calculated above in. You will get $n_{O_2}$

vera_ewing · Answer

@JFraser Is the answer to this one 1.6 mol?

jfraser · Answer

it would be 1.6 moles of lithium, but the question asks about oxygen gas. your last conversion is upside-down