From the balanced equation 2KClO3 → 2KCl + 3O2, the coefficients are the following:
coefficient 2 in front of potassium chlorate KClO3
coefficient 2 in front of potassium chloride KCl
coefficient 3 in front of oxygen molecule O2
We got this balanced equation by identifying the number of atoms of each element that we have in the given equation KClO3 → KCl + O2.
Looking at the subscripts of each atom on the reactant side and on the product side, we have
KClO3 → KCl + O2
K=1 K=1
Cl=1 Cl=1
O=3 O=2
We can see that the oxygens are not balanced. We add a coefficient 2 to the 3 oxygen atoms on the left side and another coefficient 3 to the 2 oxygen
atoms on the right side to balance the oxygens:
2KClO3 → KCl + 3O2
The coefficient 2 in front of potassium chlorate KClO3 multiplied by the subscript 3 of the oxygen atoms on the left side indicates 6 oxygen atoms just as the coefficient 3 multiplied by the subscript 2 on the right side indicates 6 oxygen atoms.
The number of potassium K atoms and chloride Cl atoms have changed as well:
2KClO3 → KCl + 3O2
K=2 K=1
Cl=2 Cl=1
O=6 O=6
We now have two potassium K atoms and two chloride Cl atoms on the reactant side, so we add a coefficient 2 to the potassium chloride KCl on the product side:
2KClO3 → 2KCl + 3O2, which is our final balanced equation.
K=2 K=2
Cl=2 Cl=2
O=6 O=6
The potassium, chlorine, and oxygen atoms are now balanced.
Answer:75%
Explanation:
First, the balanced reaction equation must be written out clearly as a guide to solving the problem. The molar masses of H3PO4 and K3PO4 are then calculated as they will be consistently required in solving the problem. The theoretical yield is obtained from the amount of H3PO4 reacted. Since 1 mole of H3PO4 yields 1 mole of K3PO4, 0.05 moles of H3PO4 yields 0.05 moles of K3PO4. The mass of K3PO4 is produced is then the product of 0.05 and it molar mass hence the theoretical yield. The % yield is calculated as shown.
Answer:
4.74 × 10³ mg
Explanation:
Given data
- Health risk limit for chloroform in groundwater: 60.0 g/L
- Volume of the sample of groundwater: 79.0 mL = 79.0 × 10⁻³ L
The maximum mass of chloroform that there could be in the sample of groundwater to meet the standards are:
79.0 × 10⁻³ L × 60.0 g/L = 4.74 g
1 gram is equal to 10³ milligrams. Then,
4.74 g × (10³ mg/1 g) = 4.74 × 10³ mg
