Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 12.9 mL of water (d= 1.00 g/mL) and 25.1 g of SO3? How much of the limiting reagent is left over? (in grams)
Added by Nancy S.
Step 1
First, we need to find the moles of water and SO3. - Moles of water = (12.9 mL) * (1.00 g/mL) / (18.015 g/mol) = 0.716 moles - Moles of SO3 = (25.1 g) / (80.066 g/mol) = 0.313 moles Show more…
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