0 g of magnesium is burnt with 0.56 g o2 in a closed vessel
Answers
Answer:
The equation is as follows:
2Mg + O2 ------> 2MgO
(s) (g) (s)
Therefore 2 moles of Mg require 1 mole of O2 to produce 2 moles of MgO.
Calculate actual moles.
moles= mass/ molar mass
Mg= 1/24 = 0.041667
O2= 0.56/32 = 0.0175
MgO=
Mole ratio of Mg: O2 = 2:1
find mass of Mg that reacts:
that means 2 mole of Mg reacts with 1 moles of O2
If 1 = 2
Then 0.0175 = 0.0175 x 2
=0.035 moles of magnesium
mass of magnesium that reacts:
mass= moles x molar mass
= 24 x 0.035
0.84 g
That means that of the 1 gram of magnesium that reacts only 0.84 grams is used
Therefore magnesium is in excess and by 0.16
Answer:
The equation is as follows:
2Mg + O2 ------> 2MgO
(s) (g) (s)
Therefore 2 moles of Mg require 1 mole of O2 to produce 2 moles of MgO.
Calculate actual moles.
moles= mass/ molar mass
Mg= 1/24 = 0.041667
O2= 0.56/32 = 0.0175
MgO=
Mole ratio of Mg: O2 = 2:1
find mass of Mg that reacts:
that means 2 mole of Mg reacts with 1 moles of O2
If 1 = 2
Then 0.0175 = 0.0175 x 2
=0.035 moles of magnesium
mass of magnesium that reacts:
mass= moles x molar mass
= 24 x 0.035
0.84 g
That means that of the 1 gram of magnesium that reacts only 0.84 grams is used
Therefore magnesium is in excess and by 0.16