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A solution with 3.40 g of silver nitrate is mixed with one holding 2.34 g of sodium chloride…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A solution with of silver nitrate is mixed with one holding of sodium chloride. White silver chloride forms: . Find the limiting reagent, the mass of silver chloride, and the mass of the other reactant left over. Molar masses: , , .
Need a reminder? The key ideaShow
The coefficients of a balanced equation give the mole ratio. Mass to moles, moles by the mole ratio, then moles to mass. At STP one mole of a gas fills . The limiting reagent has the smallest moles divided by its coefficient; it is used up and fixes the product.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Moles. Silver nitrate: . Sodium chloride: .
Step 2. Limiting reagent. The ratio is , so silver nitrate, with fewer moles, runs out first.
Step 3. Silver chloride. .
Step 4. Excess. of sodium chloride is left: .
Answer: silver nitrate limits the reaction. It gives of silver chloride, and of sodium chloride is left.
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