stochiometry?

Question

in the reaction, 2Hg+O2--->2HgO
20.0 g of Hg are reacted with 5.00 g of O2. Calculate the maximum amount, in
grams, of HgO formed.

Answer 1

2 moles Hg react with 1 mole of O2

Therefore 40.0g should react with 20.0g of O2

But there is only 5.00g of O2 available, so the Hg is in excess.

You always use the reactant not in excess for these calculations.

1 mole O2 produces 2 moles HgO

Therefore the maximum HgO that can be produced is 5.00 x 2

= 10.00 g HgO

Answer 2

First convet all the weights to g-atoms Roughly, you have 0.1 g-atom of Hg and 0.16 g-atoms of O2.

From the equation, each g-atom of Hg will produce 1 mole of HgO, and 1 g-atom of O2 will produce 2 moles of HgO.

So we check to see which, if any, of the reactants limits the production of HgO. On the basis of Hg, we can produce 0.1 mole of HgO. On the basis of O2, we can produce 0.32 moles of HgO. Thus, Hg limits the reaction, and the amount of HgO= 0.1 * mole wt of HgO.

Answer 3

Stoichiometry means moles.
The ratio of reagents is 2:1
Mole mass of Hg is 201 gm/mole
O2 is 32
moles Hg = 20/200=.1
moles O2 = 5/32 = .156
Visually you can see that O2 will be left over. Only need .05M
gms of O2 = 32*.05=1.6+20=21.6gm

Answer 4

2Hg + O2 ==> 2HgO

Hg = 200.59 g/mol
O2 = 32.00 g/mol
HgO = 216.59 g/mol

20.0g Hg / 200.59 g/mol = 0.09970 mol Hg
5.00g O2 / 32.00 g/mol = 0.1563 mol O2

0.09970 mol Hg * 2 mol HgO / 2 mol Hg = 0.09970 mol HgO
0.1563 mol O3 * 2 mol HgO / 1 mol O2 = 0.3126 mol HgO

This means that Hg is the limiting reactant, thus only 0.0.997 mols of HgO is produced.

0.0997 mol HgO * 216.59 g/mol HgO = 21.59g HgO produced