Recap - Magnesium Bromide

Part of the Stoichiometry section of Coddy's Chemistry journey. Lesson 36 of 41.

Challenge

Part of the bromine made from Dead Sea brine becomes magnesium bromide:

Mg(s) + Br₂(l) → MgBr₂(s)

A reactor holds $0.15\ \mathrm{mol}$ of magnesium, nMg, and $0.10\ \mathrm{mol}$ of bromine, nBr2. Calculate the mass of MgBr₂, m_P, that forms (184 g/mol).

Try it yourself

Solve for $m_{P}$

All lessons in Stoichiometry

1The Mole

Counting Particles: the Mole Avogadro's Number Molar Mass Mass, Moles and Particles Moles in a Formula Recap - Mole Bridge

4Equations and Mole Ratios

Balancing Equations Reading an Equation in Moles Mass to Mass Gas Volumes in Reactions Reactions in Solution Precipitation Recap - Rocket Fuel

2Gases

Molar Volume of a Gas Avogadro's Law Pressure, Volume, Temperature The Ideal Gas Law Recap - Weather Balloon

5Limiting Reactant

The Limiting Reactant What Is Left Over Industrial Scale Recap - Aluminum Plant

3Solutions

Molar Concentration Preparing a Solution Ions in Solution Ion Concentrations Dilution Recap - Lab Prep

6Practice Problems

Recap - Fertilizer Plant Recap - Acid for the Plant Recap - Gases for Ammonia Recap - Ions in the Sample Recap - The Full Plant Recap - Bromide in Brine Recap - Chlorine for BromineRecap - Magnesium BromideRecap - Silver Bromide Test Recap - Balloon at Launch Recap - Balloon High Up Recap - Before It Bursts Recap - Hydrogen from Water