Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 85

Sodium (Na) reacts with hydrogen \(\left(\mathrm{H}_{2}\right)\) to form sodium hydride (NaH). A reaction mixture contains \(10.00 \mathrm{~g} \mathrm{Na}\) and \(0.0235 \mathrm{~g} \mathrm{H}_{2}\). (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(0.428 \mathrm{~g}\) of \(\mathrm{NaH}\) is recovered. What is the percent yield of the reaction?

Short Answer

Expert verified
(a) The balanced chemical equation for the reaction is: \(2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}\) (b) Hydrogen \(\left(\mathrm{H}_{2}\right)\) is the limiting reactant. (c) The theoretical yield for this reaction is 0.5568 g of sodium hydride \(\left(\mathrm{NaH}\right)\). (d) 9.4618 g of excess sodium (Na) is left over at the end of the reaction. (e) The percent yield of the reaction is approximately 76.87%.

Step by step solution

01

Write and balance the chemical equation

First, let's write the balanced chemical equation for the reaction between sodium (Na) and hydrogen (H₂) to form sodium hydride (NaH): \[2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}\]
02

Determine the limiting reactant

To determine the limiting reactant, we first need to calculate the moles of sodium and hydrogen present in the reaction mixture. Using the molar mass of Na and H₂ (22.99 g/mol and 2.02 g/mol, respectively), we can find the moles of each reactant: Moles of Na = (10.00 g Na) / (22.99 g/mol) = 0.435 mol Na Moles of H₂ = (0.0235 g H₂) / (2.02 g/mol) = 0.0116 mol H₂ Now, divide each by the stoichiometric coefficients from the balanced equation: Moles of Na ratio = (0.435 mol Na) / 2 = 0.218 Moles of H₂ ratio = (0.0116 mol H₂) / 1 = 0.0116 Since 0.0116 is the smallest value, H₂ is the limiting reactant.
03

Calculate the theoretical yield

Using the balanced chemical equation and the moles of the limiting reactant (H₂), we can calculate the theoretical yield of NaH: Moles of NaH = 2 * moles of H₂ = 2 * 0.0116 mol = 0.0232 mol NaH Now, convert moles of NaH to grams using the molar mass of NaH (22.99 + 1.01 = 24 g/mol): Theoretical yield = (0.0232 mol NaH) * (24 g/mol) = 0.5568 g NaH
04

Calculate the amount of excess reactant remaining

To calculate the amount of excess reactant remaining, we need to determine how much Na was consumed. Since 0.0116 mol H₂ was completely consumed, then 0.0232 mol Na (2 * 0.0116 mol) is consumed. Calculate the remaining amount of Na in moles: Moles of Na remaining = 0.435 mol Na (initial) - 0.0232 mol Na (consumed) = 0.4118 mol Na Now, convert the remaining moles of Na into grams using its molar mass (22.99 g/mol): Excess reactant remaining = (0.4118 mol Na) * (22.99 g/mol) = 9.4618 g Na
05

Calculate the percent yield

The percent yield can be calculated by dividing the actual yield (0.428 g NaH) by the theoretical yield (0.5568 g NaH) and multiplying by 100%: Percent yield = (0.428 g NaH / 0.5568 g NaH) * 100% ≈ 76.87%

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

A \(1.000-g\) sample of a liquid that contains only carbon and hydrogen burns in oxygen to produce \(1.284 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What are the mass percents of the elements present in this sample? (b) What is the empirical formula for this compound? (c) The molar mass of this compound is determined to be about \(71 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound? (Hint: When attempting this problem, understand that all of the carbon in the compound burned ends up as \(\mathrm{CO}_{2}\), and all of the hydrogen in the compound burned ends up as \(\mathrm{H}_{2} \mathrm{O}\). Also, there is only 1 mole of \(C\) per mole of \(\mathrm{CO}_{2}\), but there are 2 moles of \(\mathrm{H}\) per mole of \(\mathrm{H}_{2} \mathrm{O} .\) )

How many molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\), would be in a tablet that contained \(250 \mathrm{mg}\) of aspirin? How many atoms of carbon would be in the aspirin in that tablet?

Consider the unbalanced chemical equation \(\mathrm{CaCN}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3}+\mathrm{NH}_{3}\) If you began the reaction with \(5.65 \mathrm{~g}\) of \(\mathrm{CaCN}_{2}\) and \(12.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), how many grams of \(\mathrm{NH}_{3}\) would be produced if the reaction had an \(86.0 \%\) yield?

Determine the mass percent of each element in magnesium phosphate.

How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would you need to get \(5.00 \times 10^{30}\) carbon atoms?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear 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