Question

consider the following chemical reaction: 4 al + 3 o₂ -------> 2 al₂o₃. if 12.85 g of al reacts completely with oxygen and 18.42 g of aluminum oxide is obtained, what is the percent yield of the reaction? 24.28% 65.91% 75.86% 69.76% question 3 1 pts a 15 - g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride: 2li + f₂ → 2lif. after the reaction is complete, what will be present? 0.789 moles lithium fluoride and 1.37 moles lithium 2.16 moles lithium fluoride only none of these 0.789 moles lithium fluoride only 2.16 moles lithium fluoride and 0.395 moles fluorine

Explanation:

Step1: Calculate moles of Al

The molar mass of Al is approximately 26.98 g/mol. Moles of Al = $\frac{mass}{molar\ mass}=\frac{12.85\ g}{26.98\ g/mol}\approx0.476\ mol$

Step2: Determine theoretical yield of $Al_2O_3$

From the balanced - equation $4Al + 3O_2
ightarrow2Al_2O_3$, the mole ratio of $Al$ to $Al_2O_3$ is 4:2 or 2:1. So moles of $Al_2O_3$ produced theoretically = $\frac{0.476\ mol}{2}=0.238\ mol$
The molar mass of $Al_2O_3$ is $2\times26.98\ g/mol + 3\times16.00\ g/mol=101.96\ g/mol$
Theoretical mass of $Al_2O_3$ = $moles\times molar\ mass=0.238\ mol\times101.96\ g/mol\approx24.276\ g$

Step3: Calculate percent yield

Percent yield = $\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{18.42\ g}{24.276\ g}\times100\%\approx75.86\%$

for Question 3:

Step1: Calculate moles of Li and $F_2$

The molar mass of Li is 6.94 g/mol, so moles of Li = $\frac{15\ g}{6.94\ g/mol}\approx2.16\ mol$
The molar mass of $F_2$ is $2\times19.00\ g/mol = 38.00\ g/mol$, so moles of $F_2$ = $\frac{15\ g}{38.00\ g/mol}\approx0.395\ mol$

Step2: Determine limiting reactant

From the balanced - equation $2Li+F_2
ightarrow2LiF$, the mole ratio of Li to $F_2$ is 2:1.
For 0.395 mol of $F_2$, the moles of Li required = $2\times0.395\ mol = 0.79\ mol$
Since we have 2.16 mol of Li and only 0.79 mol is required to react with 0.395 mol of $F_2$, $F_2$ is the limiting reactant.

Step3: Calculate moles of LiF produced and remaining substances

Moles of LiF produced = $2\times0.395\ mol = 0.79\ mol\approx0.789\ mol$ (due to rounding in previous steps)
Moles of Li remaining = $2.16\ mol-0.79\ mol = 1.37\ mol$

Answer:

C. 75.86%