Certainly! Let's go through this step-by-step:
1. Balanced Chemical Equation:
The given balanced chemical equation is:
[tex]\[
Na _2 SiO _3(s) + 8 HF (aq) \rightarrow H _2 SiF _6(aq) + 2 NaF (aq) + 3 H _2 O (l)
\][/tex]
2. Determine the Mole Ratio:
According to the equation, it takes `8 moles of HF` to produce `2 moles of NaF`. Therefore, we can set up a proportion to figure out how many moles of NaF are produced per mole of HF:
[tex]\[
\text{Ratio of HF to NaF} = \frac{2 \text{ moles of NaF}}{8 \text{ moles of HF}} = \frac{1}{4}
\][/tex]
3. Calculate Moles of NaF:
If we start with `0.5 moles of HF`, the moles of NaF produced can be calculated with the established ratio:
[tex]\[
\text{Moles of NaF} = \left(\frac{1}{4}\right) \times 0.5 \text{ moles of HF} = 0.125 \text{ moles of NaF}
\][/tex]
4. Molar Mass of NaF:
Next, we need to determine the molar mass of NaF.
- Sodium (Na) has an atomic mass of approximately `22.99 g/mol`.
- Fluorine (F) has an atomic mass of approximately `18.998 g/mol`.
Therefore, the molar mass of NaF is:
[tex]\[
\text{Molar Mass of NaF} = 22.99 + 18.998 = 41.988 \text{ g/mol}
\][/tex]
5. Calculate Mass of NaF Produced:
Knowing the moles of NaF and its molar mass, we can calculate the mass of NaF produced:
[tex]\[
\text{Mass of NaF} = \text{Moles of NaF} \times \text{Molar Mass of NaF} = 0.125 \text{ moles} \times 41.988 \text{ g/mol} = 5.2485 \text{ grams}
\][/tex]
Therefore, when `0.5 moles of HF` reacts with excess [tex]\( Na _2 SiO _3 \)[/tex], the mass of NaF formed is `5.2485 grams`.
