Question

If 5.738 grams of AgNO3 is mixed with 4.115 grams of BaCl2 and allowed to react according to the balanced equation: BaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ba(NO3)2(aq) What is the limiting reagent? BaCl2AgNO3 How many grams of AgCl could be produced? grams AgCl What mass, in grams, of the excess reagent will remain? grams of excess reagent

Answer 1

Limiting reagent: AgNO3

grams AgCl : 2.44 g AgCl

grams of excess reagent remain: 0.62 g BaCl2

Step-by-step explanation:

1. Change grams to mol:

AgNO3:

5.738g x (1mol/169.87g) = 0.034 mol AgNO3

BaCl2:

4.115g x (1 mol/208.23g) = 0.020 mol BaCl2

2. Limiting reagent:

AgNO3:

0.034 mol AgNO3 x (1 mol BaCL2/ 2mol AgNO3) = 0.017 mol BaCl2

BaCl2:

0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3

Limiting reagent: AgNO3

3. Grams of AgCl produced:

Using the limiting reagent:

0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl

4. Change mol to grams:

0.017 mol AgCl x ( 143.32 g AgCl /1mol AgCl) =2.44 g AgCl

5. Grams of the excess reagent:

0.034 mol AgNO3 x (1 mol BaCl2 / 2 mol AgNO3) = 0.017 mol BaCl2

0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2

0.003 mol BaCl2 x ( 208.23 g BaCl2 / 1 mol BaCl2) = 0.62 g BaCl2