Potassium chlorate is 75% pure 48 gram of oxygen would be produced from how much potassium chlorate
2KClO3 >> 2KCl +3O2
so one gets three moles of O2 for each two moles of the chlorate.
moles of O2=48/32=1.5
so we multiply that by the mole ratio: 1.5*2/3 to get the moles of the chlorate; but it is not pure, so
moles of impure chorate= 1/.75 * 1.5*2/3=4*3*2/3*3*4= 1.3333 moles
mass of that is 1.333*123 grams
To determine how much oxygen would be produced from a given amount of potassium chlorate, we need to calculate the amount of potassium chlorate needed based on its purity (75%). Here's how you can do that:
Step 1: Convert the percentage purity to a decimal value.
Purity = 75% = 75/100 = 0.75
Step 2: Calculate the amount of potassium chlorate needed to produce 48 grams of oxygen.
Let "x" be the amount of potassium chlorate in grams.
0.75x = 48
Step 3: Solve the equation for "x" to find the amount of potassium chlorate.
Divide both sides of the equation by 0.75:
x = 48 / 0.75
x = 64 grams
Therefore, you would need 64 grams of potassium chlorate (with a purity of 75%) to produce 48 grams of oxygen.