Problem: How many grams of iron(III) sulfide form when 62.0 mL of 0.135 M iron(III) chloride reacts with 45.0 mL of 0.285 M calcium sulfide?

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For this problem, we have to calculate the mass of iron (III) sulfide formed when 62.0 mL of 0.135 M iron(III) chloride reacts with 45.0 mL of 0.285 M calcium sulfide

Step 1. Write the balanced equation

calcium sulfide + iron (III) chloride → iron (III) sulfide + calcium chloride

CaS + FeCl→ Fe2S3  + CaCl2

Balancing the equation:

3 CaS + 2 FeCl→ Fe2S + 3 CaCl2 

Fe2S     2 Fe × 55.85 g/mol Fe = 111.7 g/mol
               3 S × 32.06 g/mol S = 96.18 g/mol
                               Sum 
207.88 g/mol

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Problem Details

How many grams of iron(III) sulfide form when 62.0 mL of 0.135 M iron(III) chloride reacts with 45.0 mL of 0.285 M calcium sulfide?

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry. Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems.

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Based on our data, we think this problem is relevant for Professor Stankus' class at UIW.

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.