Answer:
3.01 × 10²³ atoms Ne
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify</em>
[Given] 10.1 g Ne
[Solve] atoms Ne
<u>Step 2: Identify Conversions</u>
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
[PT] Molar Mass of Ne: 20.18 g/mol\
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
3.01398 × 10²³ atoms Ne ≈ 3.01 × 10²³ atoms Ne
Reactant C is the limiting reactant in this scenario.
Explanation:
The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.
Balanced chemical reaction is:
A + 2B + 3C → 2D + E
number of moles
A = 0.50 mole
B = 0.60 moles
C = 0.90 moles
Taking A as the reactant
1 mole of A reacted to form 2 moles of D
0.50 moles of A will produce
= 
thus 0.50 moles of A will produce 1 mole of D
Taking B as the reactant
2 moles of B reacted to form 2 moles of D
0.60 moles of B reacted to form x moles of D
= 
x = 2 moles of D is produced.
Taking C as the reactant:
3 moles of C reacted to form 2 moles of D
O.9 moles of C reacted to form x moles of D
= 
= 0.60 moles of D is formed.
Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.
To
determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample
of the compound with the given composition. Then, we calculate for the number
of moles of each element. We do as follows:<span>
mass moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59
Dividing the number of moles of each element with
the smallest value, we will have the empirical formula:
</span> moles ratio
C 4.73 / 0.59 8
H 6.50 / 0.59 11
O 1.77 / 0.59 3
N 0.59 / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>