Question

How many moles of \ce{AgCl}AgClA, g, C, l will be produced from 60.0 \text{ g}60.0 g60, point, 0, start text, space, g, end text of \cu{AgNO3}AgNO3 ​ , assuming \ce{NaCl}NaCl, a, C, l is available in excess

Answer 1

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Step-by-step explanation:

How many moles of AgCl will be produced from 60.0g AgNO3 assuming NaCl is available in excess.

Step 1: Data given

Mass of AgNO3 = 60.0 grams

Molar mass AgNO3 = 169.87 g/mol

NaCl is in excess, so AgNO3 is the limiting reactant

Step 2: The balanced equation

AgNO3 + NaCl → AgCl + NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 60.0 grams / 169.87 g/mol

Moles AgNO3 = 0.353 moles

Step 4: Calculate moles AgCl

For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl