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Chapter 3: Problem 61

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

Expert verified
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

01

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.
02

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) 脳 (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?
03

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.
04

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) 脳 (1 mol NaF / 4 mol HF) = 0.125 mol NaF
05

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) 脳 (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?
06

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
07

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) 脳 (1 mol Na2SiO3 / 8 mol HF) = 0.005 mol Na2SiO3
08

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 脳 22.99) + 28.09 + (3 脳 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) 脳 (122.07 g/mol) = 0.61 g Na2SiO3 So, 0.61 grams of Na2SiO3 can react with 0.800 g of HF.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%) .\) In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As \(\mathrm{NO}\) is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called " \(\mathrm{NO}_{\mathrm{x}}^{\prime \prime}\) gases. In \(2007,\) the United States emitted an estimated 22 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

A bottling plant has 126,515 bottles with a capacity of \(355 \mathrm{~mL}\), 108,500 caps, and \(48,775 \mathrm{~L}\) of beverage. (a) How many bottles can be filled and capped? (b) How much of each item is left over? (c) Which component limits the production?

(a) Determine the chemical formula of the product formed when the metallic element aluminum combines with the nonmetallic element bromine, \(\mathrm{Br}_{2}\). Write the balanced chemical equation for the reaction. (b) What products form when a compound containing \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) is completely combusted in air? Write a balanced chemical equation for the combustion of acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l),\) in air.

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing \(3.50 \mathrm{~g}\) of sodium carbonate is mixed with one containing \(5.00 \mathrm{~g}\) of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

The koala dines exclusively on eucalyptus leaves. Its digestive system detoxifies the eucalyptus oil, a poison to other animals. The chief constituent in eucalyptus oil is a substance called eucalyptol, which contains \(77.87 \% \mathrm{C}, 11.76 \% \mathrm{H},\) and the remainder O. (a) What is the empirical formula for this substance? (b) A mass spectrum of eucalyptol shows a peak at about 154 amu. What is the molecular formula of the substance?
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