I believe this would be the solution. Remember to first convert mass of KClO3 to moles and then use the mole to mole ratio to find the moles of KCl produced, and then convert this value into grams. To solve for the mass of KCl produced.
The modern periodic law states that the physical and chemical properties of the elements are periodic functions of their atomic numbers.
Modern periodic table is based on atomic number. Thus, the properties of elements are related to their atomic number or electronic configuration. Elements are arranged according to increasing order of atomic number. Elements with similar properties occur at regular intervals in the periodic table. The cause of periodicity in properties is due to recurrence of similar outer electronic configuration at certain regular intervals.
Answer:
Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>
The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
<h3>2. Convert Moles to Grams</h3>
Use the molar mass as a ratio.
We want to convert 3.14 moles to grams, so we multiply by that value.
The units of moles of PI₃ cancel.
<h3>3. Round</h3>
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
The 2 in the ones place tells us to leave the 9.
3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>