Chemistry, 23.06.2019 01:30, sheldonwaid4278

Magnesium is the limiting reactant in this experiment. calculate the theoretical yield of mgo for each trial. trial 1: trial 2: data mass of empty crucible with lid trial 1: 26.688 trial 2: 26.681 mass of mg metal, crucible, and lid trial 1: 26.994 trial: 2 26.985 mass of mgo, crucible, and lid trial 1: 27.188 trial 2: 27.180

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answered: Eska12

The theoretical yield of MgO for Trial 1 is 0.51 grams and for Trial 2 is 0.5066 grams.

Explanation:

Theoretical yield is calculated by the amount of reactant taken initially and is determined by the mole concept whereas experimental yield is determined with the help of experiment only.

For theoretical yield we will subtract the mass of (Mg + Crucible + Lid) to the mass of (Empty crucible + Lid)

The reaction for the formation of MgO from Magnesium follows the equation:

$2Mg+O_2\rightarrow 2MgO$

Magnesium is the limiting reagent because it limits the formation of product.

For Trial 1:

$w_1=26.688g\\w_2=26.994g$

where, w_1 is the weight of empty crucible + lid and w_2 is the weight of Mg + crucible + lid.

w_2-w_1=(26.994-26.688)g=0.306g

To calculate the moles, we use the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Given mass of magnesium = 0.306 grams

Molar mass of magnesium = 24 g/mol

Putting values in above equation, we get:

$\text{Number of moles of magnesium}=\frac{0.306g}{24g/mol}=0.01275moles$

By Stoichiometry of the Reaction:

2 moles of magnesium produces 2 moles of magnesium oxide.

So, 0.01275 moles of magnesium will produce = $\frac{2}{2}\times 0.01275=0.01275moles$ of magnesium oxide.

Now, to calculate the mass of magnesium oxide, we use equation 1:

Moles of magnesium oxide = 0.01275 moles

Molar mass of magnesium oxide = 40 g/mol

Putting values in above equation, we get:

$0.01275mol=\frac{\text{Mass of magnesium oxide}}{40g/mol}=0.5g$

Therefore, the theoretical yield of the magnesium oxide in Trial 1 is 0.5 grams.

For Trial 2:

$w_1=26.681g\\w_2=26.985g$

where, w_1 is the weight of empty crucible + lid and w_2 is the weight of Mg + crucible + lid.

w_2-w_1=(26.985-26.681)g=0.304g

Using equation 1, we get:

Given mass of magnesium = 0.304 grams

Molar mass of magnesium = 24 g/mol

Putting values in above equation, we get:

$\text{Number of moles of magnesium}=\frac{0.304g}{24g/mol}=0.01266moles$

By Stoichiometry of the Reaction:

2 moles of magnesium produces 2 moles of magnesium oxide.

So, 0.01266 moles of magnesium will produce = $\frac{2}{2}\times 0.01266=0.01266moles$ of magnesium oxide.

Now, to calculate the mass of magnesium oxide, we use equation 1:

Moles of magnesium oxide = 0.01266 moles

Molar mass of magnesium oxide = 40 g/mol

Putting values in above equation, we get:

$0.01266mol=\frac{\text{Mass of magnesium oxide}}{40g/mol}=0.5064g$

Therefore, the theoretical yield of the magnesium oxide in Trial 2 is 0.5064 grams.

answered: Guest

It shows that the experiment is designed properly.

answered: Guest

idk what is it? is this a question that you know the answer or u dont

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Magnesium is the limiting reactant in this experiment. calculate the theoretical yield of mgo for ea...