How many moles of oxygen gas will be formed from 6.45 g of potassium chlorate?

Question

Owen Ambrose · Accepted Answer

number of moles in #KClO_3# #-: 2 xx 3# = ??

This is your equation: #2KCl + 3O_2 => 2KClO_3# Assume that: n = number of moles m = mass of substance M = molar mass #n = m -: M# The mass (m) of #KClO_3# has been provided for you: 6.45 grams. Now you need to find the molar mass (M) for #KClO_3#. You know that #KClO_3# is composed of K, Cl and O. If you refer to your periodic table, find the molar mass of each atom and add them up. K = 1 #xx# 39.1 = 39.1 g/mol Cl = 1 #xx# 35.5 = 35.5 g/mol O = 3 #xx# 16.0 = 48.0 g/mol Now add them all up. 39.1 + 35.5 + 48.0 = 122.6 g/mol is the molar mass (M) of #KClO_3#. Now you can find the number of moles (n) in #KClO_3#. # n = 6.45 -: 122.6 # = 0.05261011419 moles It's best not to round yet just so it won't give you inaccurate answer at the end. If you look closely in your equation, you can see that the mole ratio between #KClO_3 : O_2# is #2:3#. If 2 moles of #KClO_3# gives you 3 moles of #O_2#, then 0.05261011419 moles of #KClO_3# will gives you: [#0.05261011419 -: 2 xx 3#] = 0.0789 moles of #O_2#. Note: This answer is rounded to 3 significant figures. Sorry if my explanation is too long or not concise, but I'm still trying my best so it makes sense to you.

Olivia Anton · Answer

undefined To calculate the number of moles of oxygen gas formed from 6.45 g of potassium chlorate, you need to use the molar mass of potassium chlorate and the stoichiometry of its decomposition reaction.

The molar mass of potassium chlorate (KClO3) is approximately 122.55 g/mol.

The decomposition reaction of potassium chlorate is:
2KClO3 -> 2KCl + 3O2

From the equation, you can see that for every 2 moles of potassium chlorate decomposed, 3 moles of oxygen gas are produced.

First, find the number of moles of potassium chlorate:
moles of KClO3 = mass / molar mass
moles of KClO3 = 6.45 g / 122.55 g/mol ≈ 0.0527 mol

Using the stoichiometry of the reaction, determine the number of moles of oxygen gas produced:
moles of O2 = (moles of KClO3) × (3 moles of O2 / 2 moles of KClO3)
moles of O2 = 0.0527 mol × (3/2) ≈ 0.0791 mol

Therefore, approximately 0.0791 moles of oxygen gas will be formed from 6.45 g of potassium chlorate.