Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
The reaction between 3.214 g magnesium and 8.416 g bromine produces 9.7 g of MgBr2.
The balanced chemical reaction of magnesium and bromine is shown below.
Mg + Br2 → MgBr2
Step 1: Convert the given mass of reactants to moles of reactants
moles Mg = mass Mg / molar mass Mg
moles Mg = 3.214 g / 24.31 g/mol
moles Mg = 0.1322 mol
moles Br2 = mass Br2 / molar mass Br2
moles Br2 = 8.416 g / 159.808 g/mol
moles Br2 = 0.0527 mol
Step 2: Bromine (Br2) is determined to be the limiting reactant. Since the stoichiometric ratio between the reactants is equal to 1, the reactant with a lower number of moles would be the limiting reactant.
Step 3: Convert moles of Br2 to the mass of MgBr2. Their stoichiometric ratio is also 1, which makes the solution easier.
mol Br2 = mol MgBr2
mass MgBr2 = mol MgBr2 * molar mass MgBr2
mass MgBr2 = 0.0527 mol * 184.118 g /mol
mass MgBr2 = 9.70 g
To learn more about stoichiometric ratios, please refer to https://brainly.com/question/9413308.
#SPJ4
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
