Answer:
16.8 g of AgCl are produced
Explanation:
The reactants are: NaCl and AgNO₃
The products are: AgCl, NaNO₃
Balanced equation: NaCl(aq) + AgNO₃(aq) → NaNO₃(aq) + AgCl(s) ↓
We convert the mass of AgNO₃ to moles → 10 g / 85g/mol = 0.117 moles
Ratio is 1:1, therefore 0.117 moles of nitrate will produce 0.117 moles of AgCl.
According to stoichiormetry.
We convert the moles to mass → 0.117 mol . 143.3g /1mol = 16.8 g
Answer:
8.44 grams of silver chloride will be produced
Explanation:
Step 1: Data given
Mass of silver nitrate = 10.0 grams
Molar mass of silver nitrate = 169.87 g/mol
Molar mass of silver chloride = 143.32 g/mol
Step 2: The balanced equation
AgNO3 + NaCl → NaNO3 + AgCl
Step 3: Calculate moles AgNO3
Moles AgNO3 = mass AgNO3 / molar mass AgNO3
Moles AgNO3 = 10.0 grams / 169.87 g/mol
Moles AgNO3 = 0.0589 moles
Step 4: Calculate moles AgCl
For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol NaNO3 and 1 mol AgCl
Step 5: Calculate mass AgCl
Mass AgCl = moles AgCl * molar mass AgCl
Mass AgCl = 0.0589 moles * 143.32 g/mol
Mass AgCl = 8.44 grams
8.44 grams of silver chloride will be produced
