1. For the reaction, calculate how many moles of the product form when 1.77mol o

Question

1. For the reaction, calculate how many moles of the product form when 1.77mol of Na completely reacts. Assume that there is more than enough of the other reactant. 2Na(s)+O2(g)--->Na2O2(s)

2. For the reaction, calculate how many moles of the product form when 2.69mol of O2 completely reacts.
Assume that there is more than enough of the other reactant.
2S(s)+3O2(g)?2SO3(g)

3. For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

2HgO(s)?2Hg(l)+O2(g)

a) 2.58 gHgO

b) 6.03 gHgO

c) 1.92 kgHgO

d) 3.67 mgHgO

3. For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

2HgO(s)?2Hg(l)+O2(g)

a) 2.58 gHgO

b) 6.03 gHgO

c) 1.92 kgHgO

d) 3.67 mgHgO

Explanation / Answer

1) number of moles of product=1.77/2=0.885

2)as one mole of O2 gives 2/3 moles of SO3,2.69 moles gives

2.69x2/3=1.793moles of product

3)atomic wt of HgO=200+16=216g

a) so 2.58g=0.0119moles

so number of moles of O2=5.972X10^-3

number of grams=0.1911gm

b)

so 6.03g=0.02791moles

so number of moles of O2=0.01395

number of grams=0.446gm

c)

so 1.92kg=8.88moles

so number of moles of O2=4.44

number of grams=142.22g

d)

so 3.67mg=1.699x10^-5moles

so number of moles of O2=8.49x10^-6

number of grams=2.71x10^-4gm

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