Question

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

Explanation:

Step1: Write the balanced chemical equation

$$\text{Ca}+\text{F}_2=\text{CaF}_2$$

Step2: Calculate the moles of reactants

Molar mass of $\text{Ca}$: $M_{\text{Ca}} = 40.08\ \text{g/mol}$
Moles of $\text{Ca}$: $n_{\text{Ca}}=\frac{m_{\text{Ca}}}{M_{\text{Ca}}}=\frac{4.0\ \text{g}}{40.08\ \text{g/mol}}\approx0.1\ \text{mol}$
Molar mass of $\text{F}_2$: $M_{\text{F}_2}=38.0\ \text{g/mol}$
Moles of $\text{F}_2$: $n_{\text{F}_2}=\frac{m_{\text{F}_2}}{M_{\text{F}_2}}=\frac{3.8\ \text{g}}{38.0\ \text{g/mol}} = 0.1\ \text{mol}$

Step3: Determine the limiting reactant

From the balanced equation, the mole ratio of $\text{Ca}$ to $\text{F}_2$ is $1:1$. Both reactants have $0.1\ \text{mol}$, so neither is in excess.

Step4: Calculate the moles of $\text{CaF}_2$

From the balanced equation, $n_{\text{CaF}_2}=n_{\text{Ca}} = n_{\text{F}_2}=0.1\ \text{mol}$

Step5: Calculate the mass of $\text{CaF}_2$

Molar mass of $\text{CaF}_2$: $M_{\text{CaF}_2}=40.08+(19.0\times2)=78.08\ \text{g/mol}$
Mass of $\text{CaF}_2$: $m_{\text{CaF}_2}=n_{\text{CaF}_2}\times M_{\text{CaF}_2}=0.1\ \text{mol}\times78.08\ \text{g/mol}=7.8\ \text{g}$

Answer:

$7.8\ \text{g}$