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Hydrofluoric acid, HF(aq), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the HF(aq). Sodium silicate (Na2SiO3), for example, reacts as follows: Na2SiO3(s)+8HF(aq)→H2SiF6(aq)+2NaF(aq)+3H2O(l)?A)How many moles of HF are needed to react with 0.260 mol of Na2SiO3?B) How many grams of NaF form when 0.600 mol of HF reacts with excess Na2SiO3?C)How many grams of Na2SiO3 can react with 0.900 g of HF?

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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]

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