A 3.214-g sample of magnesium reacts with 8.416g of bromine. The only product is magnesium bromide. If 1.934g of magnesium is left unreacted, how much magnesium bromide is formed?
-Sorry, I've completely forgot how to do this..please help
Write the equation and balance it.
Mg + Br2 ==> MgBr2
Convert 3.214 g Mg to moles. moles = grams/molar mass
Convert 8.416 g bromine to moles using th same procedure.
Since you know bromine is the limiting reagent (since Mg is in excess), work this as in any stoichiometry problem.
Convert the 1.934 g Mg to moles and subtract from moles you had to start with. That will give you the moles Mg that reacted. Then convert that to moles MgBr2 using the coefficients in the balanced equation.
moles Mg used x (1 mole MgBr2/1 mole Mg) = moles Mg used x 1 = moles MgBr2 produced.
Then g MgBr2 = moles MgBr2 x molar mass MgBr2.
To solve this problem, you can use the concept of stoichiometry. Here's how you can calculate the amount of magnesium bromide formed:
1. Calculate the amount of magnesium that reacted:
Amount of magnesium reacted = Initial amount of magnesium - Final amount of magnesium
Amount of magnesium reacted = 3.214 g - 1.934 g
Amount of magnesium reacted = 1.280 g
2. Determine the limiting reactant:
To find the limiting reactant, you need to compare the amounts of each reactant used relative to their stoichiometric coefficients. The balanced equation for the reaction is:
Mg + Br2 → MgBr2
The stoichiometric coefficients for magnesium and bromine are 1 and 1 respectively. Comparing the amounts used, we have:
Amount of magnesium used = 3.214 g - 1.934 g = 1.280 g
Amount of bromine used = 8.416 g
We can see that magnesium is the limiting reactant because less magnesium was used compared to the amount of bromine.
3. Calculate the amount of magnesium bromide formed:
Since magnesium is the limiting reactant, the amount of magnesium bromide formed will be equal to the amount of magnesium used.
Amount of magnesium bromide formed = Amount of magnesium used
Amount of magnesium bromide formed = 1.280 g
Therefore, 1.280 grams of magnesium bromide is formed.
No problem! I can help you solve this problem step-by-step.
To determine the amount of magnesium bromide formed, we need to calculate the amount of magnesium that reacted. Once we have that, we can use stoichiometry to find the amount of magnesium bromide using the balanced chemical equation.
First, let's calculate the amount of magnesium that reacted.
1. Start by subtracting the mass of unreacted magnesium from the initial mass of magnesium:
Mass of reacted magnesium = Initial mass of magnesium - Mass of unreacted magnesium
= 3.214 g - 1.934 g
= 1.280 g
Next, we need to use the balanced chemical equation for the reaction between magnesium and bromine:
Mg + Br₂ → MgBr₂
From the equation, we can see that 1 mole of magnesium reacts with 1 mole of bromine to form 1 mole of magnesium bromide. To determine the molar ratio, we need to find the molar masses of magnesium and magnesium bromide.
The atomic mass of magnesium (Mg) is approximately 24.31 g/mol.
The molar mass of magnesium bromide (MgBr₂) is the sum of the atomic masses of magnesium and bromine:
MgBr₂ = 24.31 g/mol + 2 × 79.90 g/mol = 184.11 g/mol
Now, let's convert the mass of reacted magnesium to moles, using the molar mass of magnesium:
Moles of reacted magnesium = Mass of reacted magnesium / Molar mass of magnesium
= 1.280 g / 24.31 g/mol
≈ 0.0527 mol
Since the stoichiometry of the reaction is 1:1, this means that 0.0527 mol of magnesium bromide is formed.
Finally, to find the mass of magnesium bromide, we can multiply the moles of magnesium bromide by its molar mass:
Mass of magnesium bromide = Moles of magnesium bromide × Molar mass of magnesium bromide
= 0.0527 mol × 184.11 g/mol
≈ 9.70 g
Therefore, approximately 9.70 grams of magnesium bromide are formed in this reaction.