Question

how much caf2 would form if 4.0 grams of ca reacts with 3.8 grams of f2?

reaction	reactant(s)	product(s)
mass	4.0 g	3.8 g	?

Explanation:

Step1: Write the balanced chemical equation

$Ca + F_2=CaF_2$

Step2: Calculate the molar masses

The molar mass of $Ca$ is $M_{Ca}=40.08\ g/mol$, the molar mass of $F_2$ is $M_{F_2}=2\times19.00 = 38.00\ g/mol$, and the molar mass of $CaF_2$ is $M_{CaF_2}=40.08 + 2\times19.00=78.08\ g/mol$.

Step3: Calculate the number of moles of reactants

The number of moles of $Ca$, $n_{Ca}=\frac{m_{Ca}}{M_{Ca}}=\frac{4.0\ g}{40.08\ g/mol}\approx0.1\ mol$.
The number of moles of $F_2$, $n_{F_2}=\frac{m_{F_2}}{M_{F_2}}=\frac{3.8\ g}{38.00\ g/mol}=0.1\ mol$.

Step4: Determine the limiting reactant

From the balanced - equation, the mole ratio of $Ca$ to $F_2$ is $1:1$. Since the number of moles of $Ca$ and $F_2$ are equal, neither is in excess.

Step5: Calculate the mass of $CaF_2$ formed

According to the stoichiometry of the reaction, 1 mole of $Ca$ reacts with 1 mole of $F_2$ to form 1 mole of $CaF_2$. Since 0.1 mole of $Ca$ and 0.1 mole of $F_2$ react, the number of moles of $CaF_2$ formed is $n_{CaF_2}=0.1\ mol$.
The mass of $CaF_2$ formed, $m_{CaF_2}=n_{CaF_2}\times M_{CaF_2}=0.1\ mol\times78.08\ g/mol = 7.808\ g\approx7.8\ g$.

Answer:

$7.8\ g$