Question

Hydrofluoric acid, HF \((a q)\), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q)\). Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right),\) for example, reacts as follows: $$ \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(a q)+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) How many moles of HF are needed to react with 0.300 mol of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

Short answer

a) 2.40 mol of HF are needed to react with 0.300 mol of Na2SiO3. b) 5.25 grams of NaF are formed when 0.500 mol of HF reacts with excess Na2SiO3. c) 0.610 grams of Na2SiO3 can react with 0.800 g of HF.

Step-by-step solution

Identify the balanced chemical equation

The chemical equation given is: Na2SiO3(s) + 8HF(aq) -> H2SiF6(aq) + 2NaF(aq) + 3H2O(l)

Use stoichiometry to determine moles of HF

We will use the stoichiometry of the balanced equation to determine the moles of HF needed to react with 0.300 mol of Na2SiO3: 1 mol Na2SiO3 needs 8 mol HF So, 0.300 mol Na2SiO3 will need: 0.300 mol Na2SiO3 × (8 mol HF / 1 mol Na2SiO3) = 2.40 mol HF #b) Calculate grams of NaF when 0.500 mol HF reacts with excess Na2SiO3#

Use stoichiometry to determine moles of NaF

We are given that 0.500 mol HF reacts with excess Na2SiO3. We will use the stoichiometry of the balanced equation to determine the moles of NaF formed: 8 mol HF produces 2 mol NaF So, 0.500 mol HF will produce: 0.500 mol HF × (2 mol NaF / 8 mol HF) = 0.125 mol NaF

Convert moles of NaF to grams

Now, we will use the molar mass of NaF (22.99 g/mol for Na + 18.998 g/mol for F = 41.988 g/mol) to convert moles of NaF to grams: 0.125 mol NaF × (41.988 g/mol) = 5.25 g NaF #c) Calculate grams of Na2SiO3 that can react with 0.800 g HF#

Convert grams of HF to moles

We will use the molar mass of HF (1.008 g/mol for H + 18.998 g/mol for F = 20.006 g/mol) to convert the given mass of HF (0.800 g) to moles: 0.800 g HF × (1 mol HF / 20.006 g/mol) = 0.0400 mol HF

Use stoichiometry to determine moles of Na2SiO3

We will use the stoichiometry of the balanced equation to determine the moles of Na2SiO3 that can react with 0.0400 mol HF: 8 mol HF reacts with 1 mol Na2SiO3 So, 0.0400 mol HF will react with: 0.0400 mol HF × (1 mol Na2SiO3 / 8 mol HF) = 0.00500 mol Na2SiO3

Convert moles of Na2SiO3 to grams

Now, we will use the molar mass of Na2SiO3 (2 × 22.99 g/mol for Na + 28.09 g/mol for Si + 3 × 15.999 g/mol for O = 122.07 g/mol) to convert moles of Na2SiO3 to grams: 0.00500 mol Na2SiO3 × (122.07 g/mol) = 0.610 g Na2SiO3 To sum up the answers: a) 2.40 mol HF are needed to react with 0.300 mol of Na2SiO3 b) 5.25 grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3 c) 0.610 grams of Na2SiO3 can react with 0.800 g of HF

Key concepts

Balanced Chemical Equation

The concept of a balanced chemical equation is essential in chemical reactions. It ensures that the number of atoms for each element is the same on both sides of the equation. This principle reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
In the given reaction, \[\mathrm{Na}_{2} \mathrm{SiO}_{3}(s) + 8 \mathrm{HF}(a q) \rightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(a q) + 2 \mathrm{NaF}(a q) + 3 \mathrm{H}_{2} \mathrm{O}(l)\], each element is accounted for on both sides.
For example: - Sodium (Na) has 2 atoms on both sides (from Na2SiO3 and 2NaF). - Silicon (Si) has 1 atom on both sides (from Na2SiO3 and H2SiF6). - Oxygen (O) has 3 atoms on both sides (from Na2SiO3 and 3H2O). - Fluorine (F) accounts for 8 atoms on both sides (from 8HF, 6 in H2SiF6, and 2 in NaF).
This balancing allows chemists to predict how substances interact and ensures the reaction adheres to natural laws.

Moles to Grams Conversion

Converting between moles and grams is crucial in chemistry for quantifying reactions. This process begins with the stoichiometric coefficients from the balanced chemical equation. Let's examine how this applies in the exercise using NaF and HF.
First, you identify the moles of a substance participating in the reaction, typically from the balanced equation. For instance, when 0.500 moles of HF are used, we determine the resulting moles of NaF using stoichiometry: 2 moles of NaF form from 8 moles of HF. Therefore, \(0.500\,\text{mol HF} \times \frac{2\,\text{mol NaF}}{8\,\text{mol HF}} = 0.125\,\text{mol NaF}\).
Second, utilize the molar mass of the compound to complete the conversion. NaF has a molar mass of 41.988 g/mol, calculated by adding the atomic masses of sodium (Na) and fluorine (F). Thus, \(0.125\,\text{mol NaF} \times 41.988\,\text{g/mol} = 5.25\,\text{g NaF}\).
This conversion ensures the amount of product formed is quantifiable in lab conditions.

Molar Mass Calculations

Molar mass is essential for converting between grams and moles, as it tells us the mass of one mole of a substance. To calculate molar mass, sum the atomic masses of all atoms in a formula found on the periodic table, expressed in g/mol.
For example, let's calculate the molar mass of Na2SiO3 in this exercise:

• Sodium (Na) has an atomic mass of about 22.99 g/mol. Two atoms of Na contribute: 2 x 22.99 = 45.98 g/mol.
• Silicon (Si) has an atomic mass of roughly 28.09 g/mol.
• Oxygen (O) has an atomic mass of about 15.999 g/mol. For three atoms, it’s: 3 x 15.999 = 47.997 g/mol.

Sum them to find the molar mass of Na2SiO3: 45.98 + 28.09 + 47.997 = 122.07 g/mol.
Similar calculations are done for other reactants or products in a reaction, like HF or NaF, to assist in further stoichiometric calculations.
This knowledge bridges the gap between atomic and macro scales in chemical reactions, making it pivotal for lab work and industry applications.