Calculate the mass of the product magnesium, when 6.40g of magnesium with 1.32g of oxygen were reactants.

3.32 g MgO

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.  

M_r:        24.30  32.00    40.30

               2Mg   + O₂ ⟶ 2MgO

Mass/g:  6.40      1.32

Step 2. Calculate the moles of each reactant

Moles of Mg = 6.40 × 1/24.30

Moles of Mg= 0.2634 mol Mg

Moles of O₂ = 1.32 × 1/32.00

Moles of O₂ = 0.041 25 mol O₂  

Step 3. Identify the limiting reactant  

Calculate the moles of MgO we can obtain from each reactant.  

From Mg :

The molar ratio is 2 mol MgO:2 mol Mg

Moles of MgO = 0.2634 × 2/2

Moles of MgO = 0.2634 mol MgO

From O₂:

The molar ratio is 2 mol MgO:1 mol O₂

Moles of MgO = 0.041 25 × 2/1

Moles of MgO = 0.082 50 mol MgO

The limiting reactant is O₂ because it gives the smaller amount of MgO.

Step 4. Calculate the mass of MgO that you can obtain from O₂.

Mass of MgO = 0.082 50 × 40.30/1

Mass of MgO = 3.32 g MgO

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24