Mercury metal react beside oxygen to form the bright ginger, mercury(II) oxide. 2Hg(l) + 02(g) ->2HgO(s)?

if 200 g of Hg and 30.0 g of O2 are mixed (Hg is the limiting reagent)

how many grams of HgO can be formed from this mixture?
Answers:
2Hg + O2 = 2HgO

2. Convert known masses of the molecules to moles using molecular immensity.

Hg: 2.x102 g / 200.59 (g/mol) = 1. moles
O2: 30 g / 31.9988 (g/mol) = 0.9 moles

3. Use the balanced equation to set up the appropriate mole ratios.

2 mol (Hg): 1 mol (O2): 2 mol (HgO)

4. Use the mole ratio to calculate the number of moles of the desired reactant or product. If the mass or the mole of only one species contained by the equation is known, use it to calculate the rest according to the coefficients (mole ratios) of the equation. Otherwise, find the limiting reagent and use it to add the rest.

In the equation, the moles of following molecules are known from input and their reactive qualities are:

Hg: 1. moles / 2 = 0.5 moles
O2: 0.9 moles / 1 = 0.9 moles

Therefore, the Limit reagent (the molecule near minimum reactive quantity in moles) is: Hg, multiply it by the corresponding ratio to dig up the moles for other species.

HgO: 1. moles

5. Convert from moles back to grams using molecular weights.
HgO: 1. moles x 216.59 (g/mol) = 2.x102 g
2Hg(l) + 02(g) ->2HgO(s)

First calculate the molar loads of reactants and products:

2*200 + 2*16 -> 2*(200+16)

400 + 32 -> 432

400 g of Hg produces 432g of HgO

Now use direct proportion:

200g Hg produces 200*432/400 = 216g HgO
Moles Hg = 200 g / 200.59 g/mol =0.997
Moles O2 = 30.0 g / 31.9988 g/mol =0.937
the ratio between O2 and HgO is 1 : 2
moles HgO produced = 0.937 x 2 =1.87
Mass HgO = 1.87 mol x 216.5894 g/mol =405 g

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