This is a limiting reagent (LR) problem because amounts are given for BOTH reactants. You need to find the arrow key and use it; otherwise we don't know where the reactants end and the products start.
2Mg(s) + O2(g) 2MgO(s)
mols Mg = grams/molar mass
mols O2 = grams/molar mass
Using the coefficients in the balanced eqwuation, convert mols Mg to mols MgO.
Do the same and convert mols O2 to mols MgO.
It is likely that the two values for mols MgO will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller one and the reagent producing that number is the LR.
Use the smaller number and convert to grams. g MgO = mols MgO x molar mass MgO.
2Mg(s) + O2(g) 2MgO(s)
If 9.92 g of Mg reacts with 8.52 g of O2. How many grams of MgO could be produced?
I am just very confused and need an answer so that I can look back on how you got there . . .
1 answer
1 answer