60 Grams are produced due to the fact that it may not produce more due to the certain capacity that varies on the size of the cube.
Answer : The initial quantity of sodium metal used is 17.25 grams.
Solution : Given,
Volume of
gas = 8.40 L
Molar mass of Na metal = 23 g/mole
The Net balanced chemical reaction is,

At STP, 22.4 L of volume is occupied by 1 mole of
gas
so, 8.40 L of volume is occupied by =
= 0.375 moles of
gas
Now from the above reaction, we conclude that
1 mole of
gas produced by the 2 moles of Na metal
0.375 moles of
gas produced =
of Na metal
The quantity of Na metal used = Moles of Na metal × Molar mass of Na metal = 0.75 moles × 23 g/mole = 17.25 grams
Therefore, the initial quantity of sodium metal used is 17.25 grams.
<span>the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
molar mass of Na</span>₂SO₄<span> is - 142 g/mol.
1 mol of </span>Na₂SO₄<span> has a mass of 142 g.
In 1 mol of </span>Na₂SO₄<span> the mass of Na is 23 g/mol x 2 = 46 g.
Mass of Na in 1 mol of </span>Na₂SO₄ is - 46 g
mass of Na in 0.820 mol of Na₂SO₄ - 46 g /1 mol x 0.820 mol = 37.72 g.
mass of Na is 37.72 g
<h3>
Answer:</h3>
2100 g Fe₂(SO₄)₃
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Atomic Structure</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
5.26 mol Fe₂(SO₄)₃
<u>Step 2: Identify Conversions</u>
Molar Mass of Fe - 55.85 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Fe₂(SO₄)₃ - 2(55.85) + 3(32.07) + 12(16.00) = 399.91 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2103.53 g Fe₂(SO₄)₃ ≈ 2100 g Fe₂(SO₄)₃