Answer:
[tex]\boxed{\textbf{A) 2.08 mol HF; B) 6.30 g NaF; C) 0.686 g Na$_{2}$SiO$_{3}$}}[/tex]
Explanation:
1. Moles of HF
We know we will need an equation with moles, so let’s gather all the information in one place.
Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2NaF + 3H₂O
n/mol: 0.260
The molar ratio is 8 mol HF:1 mol Na₂SiO₃
[tex]\text{Moles of HF} = \text{0.260 mol H$_{2}$SiO$_{3}$} \times \dfrac{\text{8 mol HF}}{\text{1 mol Na$_{2}$SiO$_{3}$}} = \textbf{2.08 mol HF}\\\\\text{You need $\boxed{\textbf{2.08 mol HF}}$}[/tex]
2. Mass of NaF
M_r: 122.06
Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2NaF + 3H₂O
m/g: 0.600
The molar ratio is 2 mol NaF:8 mol HF
[tex]\text{Moles of HF} = \text{0.600 mol HF} \times \dfrac{\text{2 mol NaF}}{\text{8 mol HF}} = \text{0.150 mol NaF}\\\\ \text{Mass of NaF} = \text{0.150 mol NaF} \times \dfrac{\text{41.99 g NaF}}{\text{1 mol NaF}} = \textbf{6.30 g NaF}\\\\ \text{The reaction will form $\boxed{\textbf{6.30 g NaF}}$}[/tex]
3. Mass of Na₂SiO₃
M_r: 122.06 20.01
Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2NaF + 3H₂O
m/g: 0.900
[tex]\text{Moles of HF} = \text{0.900 g HF} \times \dfrac{\text{1 mol HF}}{\text{20.01 g HF}} = \text{0.044 98 mol HF}[/tex]
The molar ratio is 1 mol Na₂SiO₃:8 mol HF
[tex]\text{Moles of Na$_{2}$SiO$_{3}$} = \text{0.044 98 mol HF} \times \dfrac{\text{1 mol Na$_{2}$SiO$_{3}$}}{\text{8 mol Na$_{2}$SiO$_{3}$}} = 5.622 \times 10^{-3}\text{ mol Na$_{2}$SiO$_{3}$}\\\\ \text{Mass of Na$_{2}$SiO$_{3}$} = 5.622 \times 10^{-3}\text{ mol Na$_{2}$SiO$_{3}$} \times \dfrac{\text{122.06 g Na$_{2}$SiO$_{3}$}}{\text{1 mol Na$_{2}$SiO$_{3}$}} = \textbf{0.686 g Na$_{2}$SiO$_{3}$}\\\\ \boxed{\textbf{0.686 g Na$_{2}$SiO$_{3}$}}\text{ will react}[/tex]
