The equation is balanced because the reactants are same as its products. so answer is B.
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Answer:
0.03697 mol Al₂(SO₄)₃
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
12.65 g Al₂(SO₄)₃
<u>Step 2: Identify Conversions</u>
Molar Mass of Al - 26.98 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Al₂(SO₄)₃ - 2(26.98) + 3(32.07) + 12(16.00) = 342.17 g/mol
<u>Step 3: Convert</u>
<u />
= 0.03697 mol Al₂(SO₄)₃
<u>Step 4: Check</u>
<em>We are given 4 sig figs. Follow sig fig rules and round.</em>
We already have 4 sig figs in the final answer, so no need to round.
Answer:
0.190L of hydrogen may be produced by the reaction.
Explanation:
Our reaction is:
3Mg + 2H₃PO₄ → Mg₃(PO₄)₂ + 3H₂
We need to determine the limting reactant. Let's find out the moles of each:
5.159×10²¹ atoms . 1 mol / 6.02×10²³ atoms = 0.00857 moles of Mg
55.23 g . 1 mol / 97.97 g = 0.563 moles of acid
2 moles of acid react to 3 moles of Mg
0.563 moles of acid may react to: (0.563 . 3) /2 = 0.8445 moles of Mg
Definetely the limting reactant is Mg.
As ratio is 3:3, 3 moles of Mg can produce 3 moles of hydrogen
Then, 0.00857 moles of Mg must produce 0.00857 moles of H₂
At STP, 1 mol of any gas occupies 22.4L
0.00857 mol . 22.4L / 1mol = 0.190L