Question

how much mgbr2 would form if 1.6 grams of mg reacts with 10.5 grams of br2?

reaction	reactant(s)	product(s)
mass	1.6 g	10.5 g	?

Explanation:

Step1: Calculate moles of Mg

The molar mass of Mg is approximately 24.31 g/mol. Using the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass and $M$ is the molar - mass. For Mg, $n_{Mg}=\frac{1.6\ g}{24.31\ g/mol}\approx0.066\ mol$.

Step2: Calculate moles of Br₂

The molar mass of Br₂ is approximately 159.81 g/mol. Using the formula $n = \frac{m}{M}$, for Br₂, $n_{Br_2}=\frac{10.5\ g}{159.81\ g/mol}\approx0.066\ mol$.

Step3: Determine the limiting reactant

The balanced chemical equation for the reaction is $Mg + Br_2
ightarrow MgBr_2$. The mole - ratio of Mg to Br₂ is 1:1. Since the moles of Mg and Br₂ are approximately equal, we can use either one to calculate the moles of MgBr₂.

Step4: Calculate moles of MgBr₂

From the balanced equation, 1 mole of Mg reacts to form 1 mole of MgBr₂. So, $n_{MgBr_2}=n_{Mg}=n_{Br_2}\approx0.066\ mol$.

Step5: Calculate mass of MgBr₂

The molar mass of MgBr₂ is $M_{MgBr_2}=24.31\ g/mol + 2\times79.90\ g/mol=24.31\ g/mol + 159.80\ g/mol = 184.11\ g/mol$. Using the formula $m = n\times M$, $m_{MgBr_2}=0.066\ mol\times184.11\ g/mol\approx12.25\ g$.

Answer:

12.25 g