(2) 3 Cu + 8 HNO3 = 3 Cu(NO3)2+ 4 H2O + 2 NO.

(3) NO + O = NO2.

(4) SO2 + H2O + NO2 =H2SO4 + NO.

(4) comes from combining the gaseous products in (1), (2), (3).
In (3), NO takes an atom of O from the air, becoming NO2, and at
once gives it up, to the H2SO3 (H2O + SO2), making H2SO4, and
again goes through the same operation of taking up O and passing
it along. NO is thus called a carrier of O. It is a reducing
agent, while NO2 is an oxidizing agent. This is a continuous
process, and very important, since it changes useless H2SO3 into
valuable H2SO4. If exposed to the air, H2SO3 would very slowly
take up O and become H2SO4.

Instead of the last experiment, this may be employed if
preferred: Burn a little S in a receiver. Put into an
evaporating-dish, 5 cc. HNO3, and dip a paper or piece of cloth
into it. Hang the paper in the receiver of SO2, letting no HNO3
drop from it. Continue this operation till a small quantity of
liquid is found in the bottle. The fumes show that HNO3 has lost
O. 2 HNO3 + SO2 = H2SO4 + 2 NO2.

91. Tests for H2SO4.

Experiment 56.--(1) Test the liquid with litmus. (2) Transfer it
to a t.t., and add an equal volume of BaCl2 solution. H2SO4 +
BaCl2 = ? Is BaSO4 soluble? (3) Put one drop H2SO4 from the