For the reaction 2 S(s) + 3 O2(g) → 2 SO3(g)
how much SO3 will be produced from 2.0 mol
O2 and excess S?
1. Cannot be determined unless the limiting
reactant is known.
2. 0.75 mol SO3
3. 1.3 mol SO3
4. 2.0 mol SO3

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

i would say b

explanation:

but i think i would need more information than that

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.