Ben,
1.use grams of Mg to moles of MgO by using molar mass of Mg, then mole ratio.
2.same process as number 1 but with O2.
3.determine which element produced less moles of MgO by comparing answer to question 1 and 2; that is your limiting reagent.
which produced less moles of MgO2? that's your answer to which is limiting reagent
If 12.01 g of Mg metal and 6.56 g of O2 gas are reacted,
1) How many moles of MgO can be produced from the given mass of Mg?
2) How many moles of MgO can be produced from the given mass of O2?
3) From the above calculations, determine which of the reactants is the limiting reagent.
Recall the balanced reaction is:
2 Mg (s) + O2 (g) 2 MgO (s)
Enter the number of moles of MgO from 12.01 of Mg here:
Enter the number of moles of MgO from 6.56 of O2 here:
The limiting reagent is?: Mg O2
Michael G.
answered8 years ago
DrBob222
answered8 years ago
see http://www.jiskha.com/display.cgi?id=1459199032
Explain Bot
answered11 months ago
To find the number of moles of MgO that can be produced from the given mass of Mg (12.01 g), we need to use the molar mass of Mg.
1) The molar mass of Mg is 24.31 g/mol (you can find this information on the periodic table). Divide the given mass of Mg by its molar mass to find the number of moles:
Number of moles of Mg = 12.01 g / 24.31 g/mol = 0.494 moles
So, 0.494 moles of Mg can produce 0.494 moles of MgO.
To find the number of moles of MgO that can be produced from the given mass of O2 (6.56 g), we need to use the molar mass of O2.
2) The molar mass of O2 is 32.00 g/mol. Divide the given mass of O2 by its molar mass to find the number of moles:
Number of moles of O2 = 6.56 g / 32.00 g/mol = 0.205 moles
So, 0.205 moles of O2 can produce 0.205 moles of MgO.
To determine the limiting reagent, we compare the ratios of moles from Mg and O2 to the stoichiometric ratio in the balanced equation.
3) The balanced equation is: 2 Mg (s) + O2 (g) → 2 MgO (s)
According to the balanced equation, for every 2 moles of Mg reacted, we need 1 mole of O2. Let's calculate the moles of Mg needed to react completely with the given moles of O2:
Moles of Mg needed = 1/2 × 0.205 = 0.103 moles
Comparing the moles of Mg available (0.494 moles) to the moles of Mg needed (0.103 moles), we can see that the moles of Mg available is greater. Therefore, Mg is in excess and O2 is the limiting reagent.
To summarize:
1) Number of moles of MgO from 12.01 g of Mg = 0.494 moles
2) Number of moles of MgO from 6.56 g of O2 = 0.205 moles
3) The limiting reagent is O2.
1) The molar mass of Mg is 24.31 g/mol (you can find this information on the periodic table). Divide the given mass of Mg by its molar mass to find the number of moles:
Number of moles of Mg = 12.01 g / 24.31 g/mol = 0.494 moles
So, 0.494 moles of Mg can produce 0.494 moles of MgO.
To find the number of moles of MgO that can be produced from the given mass of O2 (6.56 g), we need to use the molar mass of O2.
2) The molar mass of O2 is 32.00 g/mol. Divide the given mass of O2 by its molar mass to find the number of moles:
Number of moles of O2 = 6.56 g / 32.00 g/mol = 0.205 moles
So, 0.205 moles of O2 can produce 0.205 moles of MgO.
To determine the limiting reagent, we compare the ratios of moles from Mg and O2 to the stoichiometric ratio in the balanced equation.
3) The balanced equation is: 2 Mg (s) + O2 (g) → 2 MgO (s)
According to the balanced equation, for every 2 moles of Mg reacted, we need 1 mole of O2. Let's calculate the moles of Mg needed to react completely with the given moles of O2:
Moles of Mg needed = 1/2 × 0.205 = 0.103 moles
Comparing the moles of Mg available (0.494 moles) to the moles of Mg needed (0.103 moles), we can see that the moles of Mg available is greater. Therefore, Mg is in excess and O2 is the limiting reagent.
To summarize:
1) Number of moles of MgO from 12.01 g of Mg = 0.494 moles
2) Number of moles of MgO from 6.56 g of O2 = 0.205 moles
3) The limiting reagent is O2.
3 answers