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: $$ \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) \longrightarrow $$ (a) How many moles of \(\mathrm{HF}\) 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 \(\mathrm{HF}\) ?

Short answer

(a) 2.4 moles of HF are needed to react with 0.3 moles of \(Na_2SiO_3\). (b) 5.25 g of NaF is formed when 0.5 moles of HF reacts with excess \(Na_2SiO_3\). (c) 0.61 g of \(Na_2SiO_3\) can react with 0.8 g of HF.

Step-by-step solution

Balanced Equation

The balanced equation given is: \(Na_2SiO_3(s) + 8HF(aq) \longrightarrow 2NaF(aq) + H_2SiF_6(aq) + 5H_2O(l)\)

(a) Calculate moles of HF required

From the balanced equation, we can see the stoichiometric ratio of the reactants is: \(1Na_2SiO_3: 8HF\) We need to calculate the moles of \(HF\) required to react with 0.3 moles of \(Na_2SiO_3\). To do this, we can use the stoichiometric ratio: \(moles~of~ HF = 0.3~moles~of~Na_2SiO_3 * \frac{8~moles~of~HF}{1~mole~of~Na_2SiO_3}\) Moles of HF = \(0.3 * 8 = 2.4 moles\) Hence, 2.4 moles of HF are needed to react with 0.3 moles of \(Na_2SiO_3\).

(b) Calculate grams of NaF formed

From the balanced equation, the stoichiometric ratio is: \( 8HF: 2NaF\) Given 0.5 moles of HF, we need to find the moles of \(NaF\) formed: \(moles~of~NaF = 0.5~moles~of~HF * \frac{2~moles~of~NaF}{8~moles~of~HF}\) Moles of NaF = \(0.5 * \frac{2}{8} = 0.125~moles\) Now we need to convert moles of NaF to grams: \(grams~of~NaF = 0.125 moles * (1 mole ~of~ NaF / 42.0 g)\) Grams of NaF = \(0.125 * 42.0 = 5.25~g\) Therefore, 5.25 g of NaF is formed when 0.5 moles of HF reacts with excess \(Na_2SiO_3\).

(c) Calculate grams of Na2SiO3 required

Initially, we should convert grams of HF into moles using the molar mass of HF (20.01 g/mol): \(moles~of~HF = \frac{0.8g}{20.01g/mol} \) Moles of HF = 0.040 moles Now, using the stoichiometric ratio from the balanced equation, we calculate the moles of \(Na_2SiO_3\) required to react with the given moles of HF: \(moles~of~Na_2SiO_3 = 0.040~moles~of~HF * \frac{1~mole~of~Na_2SiO_3}{8~moles~of~HF}\) Moles of \(Na_2SiO_3\) = \(0.040 * \frac{1}{8} = 0.005~moles\) Finally, we convert the moles of \(Na_2SiO_3\) to grams using its molar mass (122.08 g/mol): \(grams~of~Na_2SiO_3 = 0.005~moles * 122.08~g/mol\) Grams of \(Na_2SiO_3\) = \(0.005 * 122.08 = 0.61~g\) Therefore, 0.61 g of \(Na_2SiO_3\) can react with 0.8 g of HF.

Key concepts

Chemical Reactions

A chemical reaction occurs when substances, known as reactants, transform into new substances, known as products. These transformations are characterized by the breaking and forming of bonds between atoms. In the exercise, the chemical reaction involves Hydrofluoric acid (HF) reacting with Sodium silicate (Na\(_2\)SiO\(_3\)). The result is the formation of new products, including Sodium fluoride (NaF) and others.

• Reactants: The starting materials, in this case, HF and Na\(_2\)SiO\(_3\).
• Products: New substances formed, such as NaF and Hexafluorosilicic acid (H\(_2\)SiF\(_6\)).
• Reaction Environment: Substances might appear solid (s), liquid (l), gas (g), or aqueous (aq), indicating they are dissolved in water, as with HF in this problem.

Chemical reactions are fundamental in chemistry and are utilized to transform materials. The balanced reaction provided is necessary to understand how the reactants interact to form the specific products seen in this scenario.

Balancing Chemical Equations

Balancing chemical equations is crucial in stoichiometry to reflect the actual amounts of reactants and products required, adhering to the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction. For the given reaction with HF and Na\(_2\)SiO\(_3\), the balanced equation is:

• Na\(_2\)SiO\(_3\)(s) + 8HF(aq) → 2NaF(aq) + H\(_2\)SiF\(_6\)(aq) + 5H\(_2\)O(l)
• This equation accounts for each atom on both sides to ensure they are equal, demonstrating that all atoms from the reactants show up in the products.
• The coefficients (numbers in front of substances) are adjusted to achieve this balance, showing how many molecules or moles of each substance are involved.

Balancing ensures that calculations based on the equation will accurately predict the required amounts of each reactant and product, which is essential for solving stoichiometric problems effectively.

Moles and Molar Mass

Understanding moles and molar mass allows us to count and measure substances in chemical reactions. One mole corresponds to 6.022 \(\times\) 10\(^{23}\) units (Avogadro's number) of a particular substance, enabling chemists to work with manageable numbers.

• Moles: Calculated using the ratio from the balanced equation: how many moles of one substance are required to react with moles of another.
• Molar Mass: The mass of one mole of a substance, usually expressed in grams per mole (g/mol). For example:
  • HF has a molar mass of about 20.01 g/mol.
  • NaF has a molar mass of approximately 42.00 g/mol.
  • Na\(_2\)SiO\(_3\) holds a molar mass of about 122.08 g/mol.

Using these concepts, you can convert between the mass of a substance and the moles, aiding the calculation of the necessary amounts to engage fully in the desired reactions, just as in the exercise with HF and Na\(_2\)SiO\(_3\). This process involves using conversion factors derived from the substance's molar masses. It ensures that chemical reactions are understood in terms of proportional amounts.