Question

How many grams of silver chloride are produced from 5g of Silver Nitrate reacting with an excess of barium chloride?

Answer 1

4.22g
1. Work out the equation and balance
2AgNo3 +BaCl2 ---> 2AgCl + Ba(No3)2

2. Work out formula mass of silver nitrate (should = 169.9)

3. Calculate the number of moles by doing moles = mass / formula mass so 5 divided by 169.9 = 0.0294 moles.


4. Check ratio - here the ratio is 2:2 i.e. 2 moles of silver nitrate to 2 moles of silver chloride so the moles will be the same so the moles of silver chloride is also 0.0294

5. Work out the formula mass of AgCl (always ignore big numbers at the start when working out formula mass) = 143.4

6. Work out mass by doing equation moles = mass/formula mass so
mass = moles x formula mass
mass = 0.0294 x 143.4
mass = 4.22g (3sf or 2dp)