How many g of silver chloride will be produced by reacting 10g of silver nitrate with sodium chloride?

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

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