Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Elemental sulfur occurs as octatomic molecules, \(\mathrm{S}_{8} .\) What mass \((\mathrm{g})\) of fluorine gas is needed to react completely with \(17.8 \mathrm{~g}\) of sulfur to form sulfur hexafluoride?

Short Answer

Expert verified
63.3 g of \(\mathrm{F}_2\) is needed.

Step by step solution

01

- Write the Balanced Chemical Equation

The balanced chemical equation for the reaction between elemental sulfur \(\mathrm{S}_8\) and fluorine gas \(\mathrm{F}_2\) to form sulfur hexafluoride \(\mathrm{SF}_6\) is: \[\mathrm{S}_8 + 24\mathrm{F}_2 \rightarrow 8\mathrm{SF}_6\]
02

- Molar Mass Calculation

Calculate the molar mass of \(\mathrm{S}_8\) and \(\mathrm{F}_2\):Molar mass of \(\mathrm{S}_8\): \(8 \times 32.07\, \mathrm{g/mol} = 256.56\, \mathrm{g/mol}\) Molar mass of \(\mathrm{F}_2\): \(2 \times 19.00\, \mathrm{g/mol} = 38.00\, \mathrm{g/mol}\)
03

- Calculate Moles of Sulfur

Determine the number of moles of sulfur in \(17.8\, \mathrm{g}\) of \(\mathrm{S}_8\):i.e., \[ \text{Moles of } \mathrm{S}_8 = \frac{17.8\, \mathrm{g}}{256.56\, \mathrm{g/mol}} \approx 0.0694\, \text{moles} \]
04

- Use Stoichiometry to Find Moles of Fluorine Gas

Use the stoichiometry from the balanced equation. For every 1 mole of \(\mathrm{S}_8\), 24 moles of \(\mathrm{F}_2\) are required:i.e., \[ \text{Moles of } \mathrm{F}_2 = 0.0694 \times 24 \approx 1.6656\, \text{moles} \]
05

- Find Mass of Fluorine Gas Required

Finally, calculate the mass of \(\mathrm{F}_2\) gas needed for the reaction using its molar mass: i.e., \[ \text{Mass of } \mathrm{F}_2 = 1.6656 \times 38.00\, \mathrm{g/mol} \approx 63.3\, \mathrm{g} \]

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.

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

Key Concepts

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

balanced chemical equation
In chemistry, a balanced chemical equation represents a chemical reaction with equal numbers of each type of atom on both the reactant and product sides. Balancing equations involves adjusting coefficients before compounds to ensure mass conservation.

For the reaction between elemental sulfur and fluorine gas, the balanced chemical equation is:
\[ \mathrm{S}_8 + 24\mathrm{F}_2 \rightarrow 8\mathrm{SF}_6 \]
This equation shows that one molecule of sulfur (\( \mathrm{S}_8 \)) reacts with 24 molecules of fluorine gas (\( \mathrm{F}_2 \)). The result is eight molecules of sulfur hexafluoride (\( \mathrm{SF}_6 \)). Balancing chemical equations ensures that atoms are neither created nor destroyed during the reaction. This step is crucial for accurate stoichiometry calculations.
molar mass calculation
Molar mass is the mass of one mole of a given substance, typically in grams per mole (g/mol). It is an essential concept in stoichiometry as it allows conversion between mass and moles.

To calculate the molar mass of sulfur (\( \mathrm{S}_8 \)) and fluorine gas (\( \mathrm{F}_2 \)):
  • Molar mass of \( \mathrm{S}_8 \): Sulfur has an atomic mass of 32.07 g/mol. One molecule of \( \mathrm{S}_8 \) contains eight sulfur atoms:

  • \[ 8 \times 32.07 \text{ g/mol} = 256.56 \text{ g/mol} \]
  • Molar mass of \( \mathrm{F}_2 \): Fluorine has an atomic mass of 19.00 g/mol. One molecule of \(\mathrm{F}_2\) contains two fluorine atoms:

  • \[ 2 \times 19.00 \text{ g/mol} = 38.00 \text{ g/mol} \]
    Performing these calculations gives the molar mass of the reactants, necessary for converting mass to moles.
mole-to-mole conversion
Mole-to-mole conversion is used to relate the moles of one substance to the moles of another using the coefficients from the balanced chemical equation.

In this reaction, one mole of \( \mathrm{S}_8 \) requires 24 moles of \( \mathrm{F}_2 \) to react completely.

Based on the balanced equation:
\[ \mathrm{S}_8 + 24\mathrm{F}_2 \rightarrow 8\mathrm{SF}_6 \]
For every mole of sulfur (\( \mathrm{S}_8 \)) reacting, there are 24 moles of fluorine gas (\( \mathrm{F}_2 \)).

If you know the moles of \( \mathrm{S}_8 \) are 0.0694,
  • Multiply by the stoichiometric coefficient to find moles of \( \mathrm{F}_2 \):

  • \[ 0.0694 \text{ moles of } \mathrm{S}_8 \times 24 \text{ moles of } \mathrm{F}_2 \text{ per 1 mole of } \mathrm{S}_8 \approx 1.6656 \text{ moles of } \mathrm{F}_2 \]
    This conversion helps us understand the stoichiometric relationship between reactants in the balanced chemical equation.
mass-to-mole conversion
Mass-to-mole conversion involves calculating the number of moles from a given mass of a substance. This is important for determining reactant or product quantities in chemical reactions.

To find moles of sulfur from the given mass:\( 17.8 \text{ g of } \mathrm{S}_8 \), use the molar mass calculated earlier:
  • Molar mass of \( \mathrm{S}_8 \) = 256.56 g/mol
You can find the number of moles by dividing mass by molar mass:
\[ \text{Moles of } \mathrm{S}_8 = \frac{17.8 \text{ g}}{256.56 \text{ g/mol}} \approx 0.0694 \text{ moles} \]
By using this conversion, we can move from a measurable mass to a chemical quantity in moles. From there, we can use mole-to-mole conversion to determine how much fluorine gas is needed for the reaction.

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

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example, $$ \operatorname{SrH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \operatorname{Sr}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{2}(g) $$ You wish to calculate the mass (g) of hydrogen gas that can be prepared from \(5.70 \mathrm{~g}\) of \(\mathrm{SrH}_{2}\) and \(4.75 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{SrH}_{2} ?\) (b) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{H}_{2} \mathrm{O} ?\) (c) Which is the limiting reactant? (d) How many grams of \(\mathrm{H}_{2}\) can be produced?

Propane is widely used in liquid form as a fuel for barbecue grills and camp stoves. For \(85.5 \mathrm{~g}\) of propane, calculate (a) moles of compound; (b) grams of carbon.

Cocaine \(\left(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{O}_{4} \mathrm{~N}\right)\) is a natural substance found in \(\operatorname{coca}\) leaves, which have been used for centuries as a local anesthetic and stimulant. Illegal cocaine arrives in the United States either as the pure compound or as the hydrochloride salt \(\left(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{O}_{4} \mathrm{NHCl}\right)\). At \(25^{\circ} \mathrm{C},\) the salt is very soluble in water \((2.50 \mathrm{~kg} / \mathrm{L}),\) but cocaine is much less so \((1.70 \mathrm{~g} / \mathrm{L})\) (a) What is the maximum mass (g) of the hydrochloride salt that can dissolve in \(50.0 \mathrm{~mL}\) of water? (b) If the solution from part (a) is treated with \(\mathrm{NaOH}\), the salt is converted to cocaine. How much more water (L) is needed to dissolve it?

3.58 Write balanced equations for each of the following by inserting the correct coefficients in the blanks: (a) \(-\mathrm{Cu}(s)+\underline{\mathrm{S}}_{8}(s) \longrightarrow-\mathrm{Cu}_{2} \mathrm{~S}(s)\) (b) \(-\mathrm{P}_{4} \mathrm{O}_{10}(s)+\underline{\mathrm{H}}_{2} \mathrm{O}(l) \longrightarrow-\mathrm{H}_{3} \mathrm{PO}_{4}(l)\) (c) \(-\mathrm{B}_{2} \mathrm{O}_{3}(s)+-\mathrm{NaOH}(a q) \longrightarrow-\mathrm{Na}_{3} \mathrm{BO}_{3}(a q)+-\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(-\mathrm{CH}_{3} \mathrm{NH}_{2}(g)+\underline{\mathrm{O}_{2}}(g) \longrightarrow\) \(-\mathrm{CO}_{2}(g)+-\mathrm{H}_{2} \mathrm{O}(g)+-\mathrm{N}_{2}(g)\)

Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is produced from the reaction of carbon monoxide and hydrogen: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) $$ How many moles of methanol can be produced when \(4.5 \mathrm{~mol}\) of CO reacts with 7.2 mol of \(\mathrm{H}_{2} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

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