<h3>
Answer:</h3>
0.144 moles
<h3>
Explanation:</h3>
- The relationship between mass of a compound, number of moles and molar mass of the compound is given by;
- Number of moles = Mass ÷ Molar mass
- Molar mass is equivalent to the relative formula mass of the compound that is calculated the atomic masses of the elements making the compound.
In this case;
Our compound, KClO3 will have a molar mass of;
= 39 + 35.5 + 4(16)
= 138.5 g/mol
Mass of KClO3 is 20 g
Therefore;
Number of moles = 20 g ÷ 138.5 g/mol
= 0.144 moles
Thus, the number of moles in 20 g of KClO3 is 0.144 moles
Answer:
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.
Explanation:
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.
Answer:
30.83 M
Explanation:
17.03052 re in one mole. So, if you multiply it by 30.83, you will get 535 g of ammonia.
In fact, the detailed answer is 30.827009392549122.
Answer:
1.1 × 10²⁴ atoms Mg
General Formulas and Concepts:
<u>Atomic Structure</u>
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify</em>
[Given] 1.8 mol Mg
[Solve] atoms Mg
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.08396 × 10²⁴ atoms Mg ≈ 1.1 × 10²⁴ atoms Mg
