Calculate how many grams of MgO that can be produced from 4.58 of Mg metal and 0.99 of O2 gas.
Recall the balanced reaction is:
2 Mg (s) + O2 (g) 2 MgO (s)
And the answers to part one were:
Moles of MgO from 4.58 g Mg = 0.1884
Moles of MgO from 0.99 g O2 = 0.0619
Limiting reagent = O2
Enter the mass of MgO possible (the theoretical yield)
What have you done on this problem?
Ben, I did this far below.
http://www.jiskha.com/display.cgi?id=1459199032
To calculate the mass of MgO that can be produced (the theoretical yield), we need to determine which reactant is the limiting reagent. The limiting reagent is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
From the given information, we know that 4.58 g of Mg and 0.99 g of O2 are available. To determine the limiting reagent, we compare the moles of MgO that can be produced from each reactant.
We have already calculated the moles of MgO that can be produced from 4.58 g of Mg, which is 0.1884 moles. Now, let's calculate the moles of MgO that can be produced from 0.99 g of O2.
The molar mass of O2 is 32 g/mol. Therefore, the moles of O2 can be calculated as follows:
Moles of O2 = mass of O2 / molar mass of O2
Moles of O2 = 0.99 g / 32 g/mol ≈ 0.031 moles
Now, we compare the moles of MgO produced from Mg and O2. The balanced equation tells us that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. So, the ratio of moles of MgO to moles of O2 is 2:1.
We have 0.1884 moles of MgO from Mg and 0.031 moles of MgO from O2. Since the ratio is 2:1, it means that for every 2 moles of MgO produced, 1 mole of O2 is required. Therefore, the moles of O2 required to react with 0.1884 moles of MgO is (0.1884 moles / 2 moles of MgO per O2) = 0.0942 moles.
Comparing the moles of O2 available (0.031 moles) to the moles of O2 required (0.0942 moles), we find that the moles of O2 available are insufficient to react with all the Mg.
Since O2 is the limiting reagent, it determines the maximum amount of MgO that can be produced. To calculate the theoretical yield of MgO, we use the balanced equation to determine the mole ratio between MgO and O2.
Since every 2 moles of MgO are produced from 1 mole of O2, we can calculate the moles of MgO that can be produced from 0.031 moles of O2 as follows:
Moles of MgO = 0.031 moles of O2 * (2 moles of MgO / 1 mole of O2) = 0.062 moles
Finally, we can calculate the mass of MgO using the molar mass of MgO, which is 40.3 g/mol.
Mass of MgO = Moles of MgO * Molar mass of MgO
Mass of MgO = 0.062 moles * 40.3 g/mol ≈ 2.50 grams
Therefore, the mass of MgO that can be produced (the theoretical yield) is approximately 2.50 grams.