Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 148

Consider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Suppose a \(32.9-\mathrm{g}\) sample of \(\mathrm{CaSiO}_{3}\) is reacted with \(31.8 \mathrm{L}\) of \(\mathrm{FH}\) at \(27.0^{\circ} \mathrm{C}\) and \(1.00\) atm. Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

Short Answer

Expert verified
In conclusion, the mass of SiF4 and H2O produced in the reaction is 34.178 g and 5.915 g, respectively.

Step by step solution

01

Balance the chemical equation

We have the unbalanced chemical equation: \( \mathrm{CaSiO}_{3}(s) + \mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(aq) + \mathrm{SiF}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \) To balance this equation, we need equal numbers of each atom on both sides. This can be achieved by multiplying the reactants and products with appropriate coefficients: \( \mathrm{CaSiO}_{3}(s) + 4\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(aq) + \mathrm{SiF}_{4}(g) + 2\mathrm{H}_{2} \mathrm{O}(l) \) Now the equation is balanced.
02

Find the moles of each reactant

We can use the given mass of CaSiO_3 and the number of moles from the Ideal Gas Law to find the moles of each reactant. For CaSiO_3, we have mass = 32.9 g and molecular weight = (40.08 + 28.09 + 3 × 16.00) g/mol = 100.17 g/mol, and thus we have: Moles of CaSiO_3 = (mass) / (molecular weight) = 32.9 g / 100.17 g/mol = 0.3285 mol For HF, we have volume = 31.8 L, pressure = 1.00 atm, and temperature =27.0 °C, and we can use the Ideal Gas Law: \( PV = nRT \), where R = 0.0821 L atm / mol K First, convert the temperature from Celsius to Kelvin: T = 27.0 °C + 273.15 = 300.15 K Now we can find the moles of HF: Moles of HF = (PV) / (RT) = (1.00 atm × 31.8 L) / (0.0821 L atm / mol K × 300.15 K) = 1.3032 mol
03

Find the limiting reactant

Now we need to find the limiting reactant; it will control how much of the products can be produced. To do this, compare the mole ratios of the reactants: Mole ratio of CaSiO_3 to HF: 1:4 Actual mole ratio: 0.3285 mol CaSiO_3 / 1.3032 mol HF = 0.2519 Since the actual mole ratio is less than the required mole ratio, CaSiO_3 is the limiting reactant.
04

Calculate the mass of SiF4 and H2O produced

To find the mass of products, we need to use stoichiometry based on the balanced chemical equation and the limiting reactant: Mass of SiF4 produced = (Moles of CaSiO_3 × mol SiF4 / mol CaSiO_3 ) × Molecular weight of SiF4 = (0.3285 mol × 1 mol SiF4 / 1 mol CaSiO_3) × (28.09 + 4 × 19.00) g/mol = 0.3285 mol × 104.09 g/mol = 34.178 g Mass of H2O produced = (Moles of CaSiO_3 × mol H2O / mol CaSiO_3) × Molecular weight of H2O = (0.3285 mol × 2 mol H2O / 1 mol CaSiO_3) × (2 × 1.01+ 16.00) g/mol = 0.3285 mol × 18.02 g/mol = 5.915 g In conclusion, the mass of SiF4 and H2O produced in the reaction is 34.178 g and 5.915 g, respectively.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ethene is converted to ethane by the reaction $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \stackrel{catalyst}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at \(25.0\) atm and \(300 .^{\circ} \mathrm{C}\) with a flow rate of \(1000 .\) Umin. Hydrogen at \(25.0\) atm and \(300 .^{\circ} \mathrm{C}\) flows into the reactor at a flow rate of \(1500 .\) L/min. If \(15.0 \mathrm{kg}\) \(\mathrm{C}_{2} \mathrm{H}_{6}\) is collected per minute, what is the percent yield of the reaction?

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A \(0.8765\)-\(\mathrm{g}\) sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied \(57.2 \mathrm{mL}\) at a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of \(734\) torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is \(19.8\) torr.)

Consider separate \(1.0\) -\(\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g),\) both at \(1.00\) atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

A person accidentally swallows a drop of liquid oxygen, \(\mathbf{O}_{2}(l)\) which has a density of \(1.149 \mathrm{g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{mL},\) what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{atm}\)?

At STP, \(1.0 \mathrm{L}\) \(Br\) \(_{2}\) reacts completely with \(3.0 \mathrm{L} \mathrm{F}_{2}\), producing \(2.0 \mathrm{L}\) of a product. What is the formula of the product? (All substances are gases.)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free