According to the reaction below how many grams of zinc Zn co

According to the reaction below, how many grams of zinc, Zn, could react in 196.19 mL of 1.439 M AgNO3? Zn(s) + 2 AgNO3(aq)--2 Ag(s) + Zn(NO3)2(aq)

1.439 M AgNO3 is 1.439 mol AgNO3 / 1 L of solution

now

volume of the solution = 196.19 ml = 196.19 ml x 1 L / 1000 ml

volume of the solution = 0.19619 L

now

moles of AgNo3 = volume of solution x molarity

moles of AgNo3 = 0.19619 L x 1.439 mol / L

moles of AgNO3 = 0.28231741 mol

now

moles of Zn = 0.28231741 mol AgNO3 x 1 mol Zn / 2 mol AgNO3

moles of Zn = 0.1411587 mol

now

mass of Zn = 0.1411587 mol x 65.38 g/mol

mass of Zn = 9.23 g

9.23 grams of zinc could react