Question

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?

Answer 1

To react with 0.260 mol of Na2SiO3, 2.08 mol of HF is needed. When 0.600 mol of HF reacts with excess Na2SiO3, 24.6 g of NaF is formed. 0.686 g of Na2SiO3 can react with 0.900 g of HF.

Step-by-step explanation:

In the reaction between sodium silicate (Na2SiO3) and hydrofluoric acid (HF), the ratio of Na2SiO3 to HF is 1:8. Therefore, to react with 0.260 mol of Na2SiO3, you will need 0.260 mol x 8 mol of HF/1 mol of Na2SiO3 = 2.08 mol of HF.

To determine the number of grams of NaF formed when 0.600 mol of HF reacts with excess Na2SiO3, you need to calculate the molar mass of NaF. The molar mass of NaF is 41 g/mol. Therefore, the mass of NaF formed is 0.600 mol x 41 g/mol = 24.6 g.

To calculate the grams of Na2SiO3 that can react with 0.900 g of HF, first convert the mass of HF to moles using the molar mass of HF (20 g/mol). 0.900 g x 1 mol/20 g = 0.045 mol of HF. Since the ratio of Na2SiO3 to HF is 1:8, the moles of Na2SiO3 that can react is 0.045 mol x 1 mol of Na2SiO3/8 mol of HF = 0.00563 mol. Finally, convert the moles of Na2SiO3 to grams using the molar mass of Na2SiO3 (122 g/mol). 0.00563 mol x 122 g/mol = 0.686 g of Na2SiO3 can react with 0.900 g of HF.

Answer 2

`\boxed{\textbf{A) 2.08 mol HF; B) 6.30 g NaF; C) 0.686 g Na$_(2)$SiO$_(3)$}}`

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

`\text{Moles of HF} = \text{0.260 mol H$_(2)$SiO$_(3)$} * \frac{\text{8 mol HF}}{\text{1 mol Na$_(2)$SiO$_(3)$}} = \textbf{2.08 mol HF}\\\\\text{You need $\boxed{\textbf{2.08 mol HF}}$}`

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

`\text{Moles of HF} = \text{0.600 mol HF} * \frac{\text{2 mol NaF}}{\text{8 mol HF}} = \text{0.150 mol NaF}\\\\ \text{Mass of NaF} = \text{0.150 mol NaF} * \frac{\text{41.99 g NaF}}{\text{1 mol NaF}} = \textbf{6.30 g NaF}\\\\ \text{The reaction will form $\boxed{\textbf{6.30 g NaF}}$}`

3. Mass of Na₂SiO₃

M_r: 122.06 20.01

Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2NaF + 3H₂O

m/g: 0.900

`\text{Moles of HF} = \text{0.900 g HF} * \frac{\text{1 mol HF}}{\text{20.01 g HF}} = \text{0.044 98 mol HF}`

The molar ratio is 1 mol Na₂SiO₃:8 mol HF

`\text{Moles of Na$_(2)$SiO$_(3)$} = \text{0.044 98 mol HF} * \frac{\text{1 mol Na$_(2)$SiO$_(3)$}}{\text{8 mol Na$_(2)$SiO$_(3)$}} = 5.622 * 10^(-3)\text{ mol Na$_(2)$SiO$_(3)$}\\\\ \text{Mass of Na$_(2)$SiO$_(3)$} = 5.622 * 10^(-3)\text{ mol Na$_(2)$SiO$_(3)$} * \frac{\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}`