3.0m3 of sulfer dioxide is reacted with 2.0dm3 of oxygen . What volume of sulfur trioxide is form?

3.0 dm^3 of sulphur trioxide is formed

Explanation:

From Gay Lussac's law of combining volumes;

The equation is;

                                                  SO2(g) + 1/2O2(g) > SO3(g)

Reacting volumes                       1 vol          1/2 vol             1 vol

Initial volumes                           3.0 dm^3    2.0 dm^3       0.0 dm^3

Final volumes                            0.00 dm^3  1.5 dm^3        3.0 dm^3

Hence, 3.0 dm^3 of sulphur trioxide is formed.

the formula would be mgso4

mgso[subscript]4

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g