0.500 L of 0.325 mol/L AgNO3 solution reacts with 2.00 g of solid Zinc in a beaker 2AgNO3(aq) + Zn 🡺 2 Ag(s) + Zn(NO3)2( aq ) assuming yield is 80% what is mass of Ag produced

Answer 1

0.500 x 0.325 = 0.1625 moles Ag

2.00/65.38 g/mole = 0.0306 moles of Zn.

1Zn —-> 2Ag, so:

0.0306 moles Zn——> 0.0612 moles Ag

at 80% yield ===> 0.0612 x 0.80 = 0.04896 moles Ag

0.04896moles x 107.838g/mole = 5.241 g of Ag.