Two moles of Mg and five moles of O2 are placed in a reaction vessel, and then the Mg is ignited to produce Mg (F.W.= 40.31 g/mole). What is the actual mole ratio, the calculated mole ratio, how many moles of MgO are formed, and what is the weight of MgO formed?
You made a typo; I'm sure you wanted that to read....."the Mg is ignited to produce MgO."
The answers are
2Mg + O2 ==> 2MgO
Two mols Mg will produce two mols MgO (in an excess of oxygen).
It appears the other questions relate to an experiment conducted with Mg and O2; therefore, you are to substitute the value you found in the experiment.
When your body metabolizes amino acids, one of the final end products is urea, a water so liable compound that is removed from the body as urine. Why is urea soluble in water, when hexamide, a related compound is not?
To determine the actual mole ratio, calculated mole ratio, moles of MgO formed, and the weight of MgO formed, we need to first write out the balanced chemical equation for the reaction.
The balanced equation for the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO) is:
2 Mg + O2 -> 2 MgO
Now, let's break down the given information step-by-step:
Step 1: Calculate the actual mole ratio
In the balanced equation, the mole ratio between Mg and O2 is 2:1. This means that for every 2 moles of Mg, we need 1 mole of O2. In the given information, we have 2 moles of Mg and 5 moles of O2. However, since O2 is in excess (more than required), we need to use it all up before considering the moles of Mg.
So, the actual mole ratio is 2 moles of Mg : 1 mole of O2.
Step 2: Calculate the calculated mole ratio
To calculate the mole ratio, we need to find the number of moles of Mg and O2 by dividing their given amounts by their respective molar masses.
Number of moles of Mg = given amount of Mg (in moles)
Number of moles of MgO = molar mass of Mg (in g/mol) ÷ molar mass of MgO (in g/mol)
Given amount of Mg = 2 moles
Molar mass of Mg = 40.31 g/mol
Molar mass of MgO = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol
Number of moles of MgO = (40.31 g/mol) ÷ (40.31 g/mol)
= 1 mole
Using these values, the calculated mole ratio is 2 moles of Mg : 1 mole of MgO.
Step 3: Calculate moles of MgO formed
Since the calculated mole ratio is 2:1, we can conclude that the number of moles of MgO formed is the same as the number of moles of Mg in the reaction. Thus, 2 moles of MgO are formed.
Step 4: Calculate the weight of MgO formed
Since the molar mass of MgO is 40.31 g/mol, we can calculate the weight of MgO formed using the formula:
Weight of MgO formed = number of moles of MgO × molar mass of MgO
Number of moles of MgO = 2 moles
Molar mass of MgO = 40.31 g/mol
Weight of MgO formed = 2 moles × 40.31 g/mol = 80.62 g
Therefore, the weight of MgO formed is 80.62 grams.
To determine the actual and calculated mole ratio, moles of reactants, moles of product formed, and weight of the product, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.
The balanced chemical equation for the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO) is:
2 Mg + O2 -> 2 MgO
Here's how to find the answers step by step:
1. Determine the actual mole ratio:
The balanced equation tells us that the ratio of moles between Mg and O2 is 2:1. Since we have 2 moles of Mg and 5 moles of O2, we can say that the actual mole ratio is 2:5.
2. Calculate the mole ratio:
To calculate the mole ratio, we need the balanced equation. From the equation, we see that the ratio of moles between Mg and O2 is 2:1. So the calculated mole ratio is also 2:5.
3. Determine the moles of MgO formed:
According to the balanced equation, for every 2 moles of Mg reacted, we obtain 2 moles of MgO. Since we have 2 moles of Mg, we will also have 2 moles of MgO.
4. Calculate the weight of MgO formed:
The formula weight (FW) of MgO is 40.31 g/mole. Since we have 2 moles of MgO, we can calculate the weight by multiplying the moles by the FW:
Weight of MgO = 2 moles * 40.31 g/mole = 80.62 g
Therefore, for the given reaction, the actual mole ratio is 2:5, the calculated mole ratio is also 2:5, 2 moles of MgO are formed, and the weight of MgO formed is 80.62 grams.