Consider these equations: 2S (s) + 3O2 (g)→2SO3 (g) , ΔH = −792 kJ 2S (s) + 2O2 (g)→2SO2 (g) , ΔH = −594 kJ 2SO2 (g) + O2 (g)→2SO3 (g) , ΔH =?
What is the missing ΔH ?
−294 kJ
−198 kJ
+198 kJ
+294 kJ

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

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

the answers are

c and c