CS2(l) + O2(g) CO2(g) + SO2(g)
Consider the above unbalanced equation. What mass of CS2 is required to react with excess oxygen to produce 985 mL of SO2 at 465 mm Hg and 54.0°C? Use molar masses with at least as many significant figures as the data given.
Answers
Answer:
Given,
325g of Na2O2 is reacted with water
Molar mass of Na2O2=78g/mol
∴ Number of moles of Na2O2 given=78g/mol325g
=4.166 moles
The reaction involved is
2Na2O2+2H2O⟶4NaOH+O2
2 moles of Na2O2 gives 1 moles of O2
∴4.166 moles of Na2O2 gives ⟶24.166 moles of O2
=2.083 moles of O2
Now,
The molar mass of CS2=76/gmol
The reaction involved id
CS2+3O2⟶CO2+2SO2
1 mole of CS2 requires 3 moles of
Explanation:
Given,
325g of Na
2
O
2
is reacted with water
Molar mass of Na
2
O
2
=78g/mol
∴ Number of moles of Na
2
O
2
given=
78g/mol
325g
=4.166 moles
The reaction involved is
2Na
2
O
2
+2H
2
O⟶4NaOH+O
2
2 moles of Na
2
O
2
gives 1 moles of O
2
∴4.166 moles of Na
2
O
2
gives ⟶
2
4.166
moles of O
2
=2.083 moles of O
2
Now,
The molar mass of CS
2
=76/gmol
The reaction involved id
CS
2
+3O
2
⟶CO
2
+2SO
2
1 mole of CS
2
requires 3 moles of O
2
Now, 2.083 moles of O
2
will react with x moles of CS
2
to produce SO
2
& CO
2
x=0.6943 moles of CS
2
∴ The mass of CS
2
which can be completely oxidized by 2.083 moles of O
2
=0.6943mol×76g/mol
=52.76g