The first-order decomposition of cyclopropane has a rate constant of 6.7 * 10-4 5-1. if the initial concentration of cyclopropane is 1.33 m, what is the concentration of cyclopropane after 644 s? oa) 0.43 m b) 0.15 m c) 0.94 m d) 0.86 m

D) 0.86 M

Explanation:

Given that:

The rate constant, k = 6.7×10⁻⁴ s⁻¹

Initial concentration [A₀] = 1.33 M

Time = 644 s

Using integrated rate law for first order kinetics as:

Where,

is the concentration at time t

So,

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b