3CuCl2 + 2Al = 2AlCl3 + 3Cu using the molar ratio between the limiting reactant (aluminum) and the product copper, determine the
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Answer:
0.282 M
General Formulas and Concepts:
<u>Chemistry - Solutions</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Molarity = moles of solute / liters of solution
Explanation:
<u>Step 1: Define</u>
5.85 g KI
0.125 L
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of I - 126.90 g/mol
Molar Mass of KI - 39.10 + 126.90 = 166 g/mol
<u>Step 3: Convert</u>
<u /> = 0.035241 mol KI
<u>Step 4: Find Molarity</u>
M = 0.035241 mol KI / 0.125 L
M = 0.281928
<u>Step 5: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.281928 M ≈ 0.282 M