Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO.
of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.
Answer:
Solid, Liquid, Gas
Hope it helps! ^^
The balanced chemical equation of the reaction described above is,
C2H6O + O2 --> H2O + C2H4O2
If we have 3.84 g of oxygen, we divide by its molar mass.
n = (3.54 g Oxygen gas) x (1 mole O2/ 32 g O2)
n = 0.11 moles O2
Using ratio and proportion,
number of moles of ethanol = (0.11 moles O2) x (1 mole C2H6)
= 0.11 moles C2H6
Then, we multiply the calculated value to its molar mass, 46 grams /mol.
mass of ethanol = (0.11 mol) x (46 grams / mol)
= <em>5.06 grams</em>
Answer:
0.03682 mL of mercury
Explanation:
We know the density of the mercury which is 13.58 g/mL
density = mass / volume
volume = mass / density
Now we can calculate the volume of 0.5 g of mercury:
volume = 0.5 / 13.58 = 0.03682 mL of mercury