how many grams of KClO3 are needed to produce 42g of 02 if the percent yield is 65.%?

Answer 1

To solve this problem,
--First convert the desired number of grams of O2 produced (42 g) into moles of O2 produced by dividing by the molar mass (32 g/mol of O2).
--Use the mole ratio from the chemical reaction to find a relationship between the moles of O2 produced and moles of KClO3 reacted to find the number of moles of KClO3 needed.
--Multiply the moles of KClO3 needed by the molar mass of KClO3 (122.55 g/mol KClO3) to find the number of grams of KClO3 needed.
--Adjust for efficiency

42 g of O2 = 1.3125 moles of O2.

The reaction which occurs is,
2KClO3 --> 3O2 + 2KCl
For every 3 moles of O2 produced, 2 moles of KClO3 must react...a ratio of 3/2.
In order to produce 1.3125 moles of O2, .875 moles of KClO3 must react.

.875 moles of KClO3 * 122.55 g/mol KClO3 = 107.23125 g
.875 moles of KClO3 has a mass of 107.23 grams.

If the efficiency of the reaction is only 65%, then divide the number of grams of KClO3 needed under idea (100% efficient) conditions by the percentage (in decimal form),
107.23 g / .65 = 164.97 g

164.97 grams of KClO3 are needed to react at 65% efficiency to produce 42 grams of O2.

Answer 2

p.c. yield= ( surely yield / Theoretical yield)* a hundred% ( sixty 5/a hundred )= forty two/x solved for x .sixty 5 x=forty two .sixty 5 = p.c. yield forty two= surely yield or produce interior the reaction x = theoretical yield =sixty 4.62g O2 based on the equation above the KClO3 is mandatory would be: sixty 4.62g O2 (1mol/32gO2)= 2.02mol O2 2.02molO2(2molKClO3/3molO2) = a million.35 mol of KClO3 * 122.55gKClO3 = one hundred sixty five.44g KClO3 notice this equivalent : sixty 4.62g O2*.sixty 5 = 42g O2

Answer 3

165 g

At 100 % yield: you get 48 g oxygen from 122.55 g of KClO3
at 65 % yield: you get 0.65*48= 31.2 g oxygen
for 42 g O2 you need 42/31.2 * 122.55 = 165 g