## You know 1 mol oxygen (32.0 grams) occupies 22.4L at STP. Convert 140 mL to grams, then,

%oxygen = (mass oxygen/mass original sample)*100 = ??%

DrBob222
answered

13 years ago

13 years ago

%oxygen = (mass oxygen/mass original sample)*100 = ??%

Explain Bot
answered

11 months ago

11 months ago

Let's consider the equation for the reaction of KClO3 decomposing into KCI and O2:

2 KClO3 -> 2 KCI + 3 O2

We know that the volume of the O2 generated at STP is 140 ml. At STP (Standard Temperature and Pressure), temperature is 273 K and pressure is 1 atm. At STP, 1 mole of any gas occupies 22.4 liters (or 22.4 * 1000 ml).

So, the number of moles of O2 generated can be calculated as follows:

moles of O2 = (volume of O2 generated / volume of 1 mole of O2 at STP)

moles of O2 = (140 ml / 22.4 * 1000 ml)

moles of O2 = 0.00625 moles

Now, according to the balanced equation, we know that 2 moles of KClO3 produce 3 moles of O2. Therefore, the number of moles of KClO3 in the mixture can be calculated as follows:

moles of KClO3 = (moles of O2 / 3) * 2

moles of KClO3 = (0.00625 moles / 3) * 2

moles of KClO3 = 0.00417 moles

Next, we need to find the moles of KCI in the mixture. Since the original mixture weighs 1.80 grams, we can use the molar mass of KCI to convert grams to moles. The molar mass of KCI is 39.10 + 35.45 = 74.55 g/mol.

moles of KCI = (mass of KCI / molar mass of KCI)

moles of KCI = (1.80 g / 74.55 g/mol)

moles of KCI = 0.02415 moles

Finally, to calculate the percent of KClO3 in the original mixture, we can use the following equation:

Percent of KClO3 = (moles of KClO3 / total moles) * 100

total moles = moles of KClO3 + moles of KCI = 0.00417 moles + 0.02415 moles = 0.02832 moles

Percent of KClO3 = (0.00417 moles / 0.02832 moles) * 100

Percent of KClO3 = 14.7%

Therefore, approximately 14.7% of the original mixture is KClO3.