The decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 °C is first order in H2O2. H2O2(aq)H2O(l) + ½ O2(g) During one experiment it was found that the H2O2 concentration dropped from 5.16×10-2 M at the beginning of the experiment to 1.95×10-2 M in 832 min. What is the value of the rate constant for the reaction at this temperature? min-1

The rate constant is 1.98×10^-3 min^-1

Explanation:

The reaction follows a first order

Rate = k[H2O2] = change in concentration of H2O2/time

Initial concentration of H2O2 = 5.16×10^-2 M

Concentration of H2O2 in 832 min = 1.95×10^-2 M

Change in concentration = 5.16×10^-2 - 1.95×10^-2 = 3.21×10^-2 M

Time = 832 min

1.95×10^-2k = 3.21×10^-2/832

k = 3.86×10^-5/1.95×10^-2 = 1.98×10^-3 min^-1

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