What is the mass of S2Cl2 if 342 grams of S8 is reacted with an excess of Cl2?
What is the percent yield of a reaction in which 240. g phosphorus trichloride reacts with excess water to form 142 g HCl and aqueous phosphorus acid?
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http://www.jiskha.com/science/chemistry/stoichiometry.html
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To determine the mass of S2Cl2 formed, we first need to balance the chemical equation for the reaction between S8 and Cl2, then use stoichiometry to calculate the mass.
The balanced chemical equation for the reaction is:
S8 + 4Cl2 → 4S2Cl2
According to the equation, 1 mole of S8 reacts with 4 moles of Cl2 to produce 4 moles of S2Cl2. The molar mass of S8 is approximately 256.52 grams/mole.
Step 1: Calculate the moles of S8 present in 342 grams of S8.
Moles of S8 = Mass of S8 / Molar mass of S8
Moles of S8 = 342 grams / 256.52 grams/mole
Moles of S8 ≈ 1.33 moles
Step 2: Use stoichiometry to determine the moles of S2Cl2 formed.
According to the balanced equation, 1 mole of S8 produces 4 moles of S2Cl2.
Moles of S2Cl2 = Moles of S8 × (4 moles of S2Cl2 / 1 mole of S8)
Moles of S2Cl2 = 1.33 moles × (4 mol S2Cl2 / 1 mol S8)
Moles of S2Cl2 ≈ 5.32 moles
Step 3: Calculate the mass of S2Cl2 formed.
Mass of S2Cl2 = Moles of S2Cl2 × Molar mass of S2Cl2
Mass of S2Cl2 = 5.32 moles × (l) grams/mole
Mass of S2Cl2 = l grams
Therefore, the mass of S2Cl2 formed is approximately equal to 5.32 grams.