Question

Hydrofluoric acid, \(\mathrm{HF}(a q),\) 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: $$ \begin{aligned} \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) & \longrightarrow \\ \mathrm{H}_{2} \mathrm{SiF}_{6}(a q) &+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ (a) How many moles of HF are needed to react with 0.300 \(\mathrm{mol}\) 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) 2.4 moles of 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) 0.61 grams of Na2SiO3 can react with 0.800 g of HF.

Step-by-step solution

Determine the mole ratio of HF to Na2SiO3

From the balanced equation: Na2SiO3(s) + 8HF(aq) -> H2SiF6(aq) + 2NaF(aq) + 3H2O(l), it can be seen that the mole ratio of HF to Na2SiO3 is 8:1.

Calculate moles of HF required

Using the mole ratio of HF to Na2SiO3, we can calculate the moles of HF needed to react with 0.300 mol of Na2SiO3. Moles of HF = (0.300 mol Na2SiO3) × (8 mol HF / 1 mol Na2SiO3) = 2.4 mol HF So, 2.4 moles of HF are needed to react with 0.300 mol of Na2SiO3. b) How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

Determine the mole ratio of NaF to HF

From the balanced equation: Na2SiO3(s) + 8HF(aq) -> H2SiF6(aq) + 2NaF(aq) + 3H2O(l), it can be seen that the mole ratio of NaF to HF is 2:8 or 1:4.

Calculate moles of NaF formed

Using the mole ratio of NaF to HF, we can calculate the moles of NaF formed when 0.500 mol of HF reacts with excess Na2SiO3. Moles of NaF = (0.500 mol HF) × (1 mol NaF / 4 mol HF) = 0.125 mol NaF

Calculate grams of NaF formed

To calculate the grams of NaF formed, we need to find the molar mass of NaF. Na has a molar mass of 22.99 g/mol and F has a molar mass of 19.00 g/mol. Molar mass of NaF = 22.99 + 19.00 = 41.99 g/mol Now, we can multiply the moles of NaF by its molar mass to get the grams of NaF. Grams of NaF = (0.125 mol NaF) × (41.99 g/mol) = 5.25 g NaF So, 5.25 grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3. c) How many grams of Na2SiO3 can react with 0.800 g of HF?

Convert grams of HF to moles

First, we need to find the molar mass of HF. H has a molar mass of 1.01 g/mol and F has a molar mass of 19.00 g/mol. Molar mass of HF = 1.01 + 19.00 = 20.01 g/mol Now, we can convert the grams of HF to moles. Moles of HF = (0.800 g HF) / (20.01 g/mol) = 0.040 moles HF

Calculate moles of Na2SiO3

Using the mole ratio of Na2SiO3 to HF, we can calculate the moles of Na2SiO3 that can react with 0.040 moles of HF. Moles of Na2SiO3 = (0.040 mol HF) × (1 mol Na2SiO3 / 8 mol HF) = 0.005 mol Na2SiO3

Calculate grams of Na2SiO3

To calculate the grams of Na2SiO3, we need to find the molar mass of Na2SiO3. Na has a molar mass of 22.99 g/mol, Si has a molar mass of 28.09 g/mol and O has a molar mass of 16.00 g/mol. Molar mass of Na2SiO3 = (2 × 22.99) + 28.09 + (3 × 16.00) = 122.07 g/mol Now, we can multiply the moles of Na2SiO3 by its molar mass to get the grams of Na2SiO3. Grams of Na2SiO3 = (0.005 mol Na2SiO3) × (122.07 g/mol) = 0.61 g Na2SiO3 So, 0.61 grams of Na2SiO3 can react with 0.800 g of HF.

Key concepts

Chemical Reactions

Chemical reactions are processes where substances are transformed into new ones with different properties. In this exercise, we're looking at a reaction between hydrofluoric acid (HF) and sodium silicate (Na\(_2\)SiO\(_3\)). This is expressed through a balanced chemical equation, which shows the reactants and the products formed. Understanding the equation is crucial as it provides insight into what happens during the reaction. Each substance in the equation has a specific role to play:

• The reactants, Na\(_2\)SiO\(_3\) and HF, combine to produce new substances.
• The products of this reaction are H\(_2\)SiF\(_6\), NaF, and H\(_2\)O. Each of these products has its own uses or implications in different contexts.

Understanding chemical reactions involves recognizing the transformation that reactants undergo, leading to new products. This is not just about mixing substances but involves energetic interactions that change molecular structures, enabling the formation of new compounds.

Molar Mass Calculations

Molar mass calculations are essential in stoichiometry as they allow us to convert between grams and moles, a key unit of measure in chemistry. The molar mass of a substance is the mass of one mole of its entities (atoms, molecules, etc.) and is often expressed in grams per mole (g/mol). For instance, for the substance sodium fluoride (NaF):

• To find the molar mass, you add together the atomic masses of sodium (22.99 g/mol) and fluorine (19.00 g/mol). This results in a molar mass of 41.99 g/mol for NaF.

Similarly, for sodium silicate (Na\(_2\)SiO\(_3\)), the calculation is slightly more complex but follows the same principle.

• Combine the molar masses of its constituent atoms (2 sodium, 1 silicon, 3 oxygen) for a total of 122.07 g/mol.

Having calculated molar masses, you can then use these values to convert between the mass of a substance and the number of moles, a critical step in solving stoichiometric problems.

Mole-to-Mole Ratios

Mole-to-mole ratios are derived from balanced chemical equations and are pivotal in predicting how much product you can get from a given amount of reactants. This ratio comes from the coefficients of the substances in the equation. For instance, in our example reaction of Na\(_2\)SiO\(_3\) and HF:

• The balanced equation shows that 1 mole of Na\(_2\)SiO\(_3\) reacts with 8 moles of HF, so the mole ratio of HF to Na\(_2\)SiO\(_3\) is 8:1.

Understanding these ratios allows you to perform calculations to find out how much of a reactant you need or how much product can be produced. In the exercise:

• To find out how many moles of HF are needed for a specific amount of Na\(_2\)SiO\(_3\), multiply the moles of Na\(_2\)SiO\(_3\) by the ratio (8 mol HF per 1 mol Na\(_2\)SiO\(_3\)).
• This gives you 2.4 moles of HF for 0.3 moles of Na\(_2\)SiO\(_3\).

These ratios and calculations are intrinsic to performing stoichiometry as they help you bridge the gap between theoretical equations and practical laboratory quantities.