For the reaction represented by the equation 2kclo3 → 2kcl + 3o2, how many grams of potassium chlorate are required to produce 160 g of oxygen? type the numeric value only.

 The grams  of potassium chlorate  that are required  to produce 160 g of oxygen  is   408.29  grams

calculation

 2 KClO₃→  2 KCl  + 3O₂

Step 1:  find the  moles of  O₂

moles  =  mass÷  molar mass

from periodic table  the  molar mass of O₂  = 16 x2 = 32 g/mol

moles  = 160 g÷ 32  g/mol =  5 moles

Step2 : use the  mole   ratio to determine the  moles of KClO₃

from equation given KClO₃ : O₂  is 2:3

therefore the v moles of KClO₃  =  5 moles x 2/3 = 3.333  moles

Step 3:  find the mass  of KClO₃

mass= moles x molar mass

from periodic table  the molar mass of KClO₃

= 39 + 35.5 + (16 x3) =122.5 g/mol

mass  = 3.333 moles x 122.5 g/mol =408.29 grams

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

idk what the question is about