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S + колёсико мыши
Непрозрачность
,
.
O + колёсико мыши
48 grams of Magnesium metal. Assume Oxygen gas in excess.
Question 1: How many grams of MgO do you expect. (Mg = 24 g/mol, MgO = 40 g/mol)?
1. Convert mass Mg -> mol Mg
48 g Mg ( 1 mol Mg / 24 g Mg) = 2 mol Mg
2. Relationship between mol Mg and mol MgO
for every 2 mol Mg reactant, you produce 2 mol MgO product
3. Find mol MgO from mol Mg
2 mol Mg ( 2 mol MgO / 2 mol Mg) = 2 mol MgO
4. Convert mol MgO to g MgO
2 mol MgO ( 40 g Mg / 1 mol MgO ) = 80 grams MgO
Magnesium metal excess. You only have 128 grams of O2 to do combustion.
Question 2: How many moles of MgO do you expect. (same mmass as before, O2 = 32 g/mol)
1. 128 grams O2 ( 1 mol O2 / 32 g O2 ) = 4 mol O2
2. Relationship between mol O2 and mol MgO
for every 1 mol O2 in, 2 mol MgO go out
3. Find mol MgO from mol O2
4 mol O2 ( 2 mol MgO / 1 mole O2) = 8 mol MgO
