In the equation #2KClO_3 - 2KCl + 3O_2#, if 5.0 g of #KClO_3# is decomposed, what volume of #O_2# is produced at STP?

Question

In the equation #2KClO_3 -> 2KCl + 3O_2#, if 5.0 g of #KClO_3# is decomposed, what volume of #O_2# is produced at STP?

Natalie Anton · Accepted Answer

#1.371L#

Find the relative mass of the molecule #KClO_3# by adding together the masses of each atom in it. Find the masses from the periodic table.

#39.01gmol^-1 + 35.45gmol^-1 + 3 * 16.00gmol^-1# # = 122.46gmol^-1#

Divide the total mass of the substance by its relative mass to find the number of moles of the substance.

#(5.00g)/(122.46gmol^-1) = 0.041mol#

There are two moles of #KClO_3# for every 3 moles of #O_2#, which we can see from the equation. Using the ratio #2:3# and inserting #0.0408mol#, we find that there must be #0.0612mol# of #O_2#.

One mole of an ideal gas at STP (standard temperature and pressure) occupies 22.4L.

Multiplying

#22.4Lmol^-1 * 0.0612mol = 1.371L#

which should be your final answer, depending on how you've rounded.

Mateo Aguilar · Answer

undefined Using stoichiometry, the molar mass of KClO₃ is 122.55 g/mol. Therefore, 5.0 g of KClO₃ is equivalent to approximately 0.041 mol. According to the balanced equation, 3 moles of O₂ are produced for every 2 moles of KClO₃ decomposed. At STP, 1 mole of any gas occupies 22.4 liters. Thus, 0.041 mol of O₂ will occupy approximately 0.055 L at STP.

In the equation #2KClO_3 -&gt; 2KCl + 3O_2#, if 5.0 g of #KClO_3# is decomposed, what volume of #O_2# is produced at STP?

In the equation #2KClO_3 -> 2KCl + 3O_2#, if 5.0 g of #KClO_3# is decomposed, what volume of #O_2# is produced at STP?