A metallic zinc tape is placed in a beaker containing 120 mL of a solution of copper (II) nitrate, Cu (NO3) 2 (aq). The copper produced has a mass of 0.813g. Find the initial concentration of the copper nitrate solution
According to the chemical reaction:
Cu(NO3)2 + Zn = Cu + Zn(NO3)2
From here, the amount of moles of copper produced in the reaction equals:
n(Cu) = m(Cu) / Mr(Cu) = 0.813 g / 63.546 = 0.0128 mol
Finally, the initial concentration of the copper nitrate solution equals:
c(Cu(NO3)2) = n(Cu(NO3)2) / V(Cu(NO3)2) = 0.0128 mol / 120 mL = 0.0128 mol / 0.120 L = 0.107 mol/L
Answer: 0.107 mol/L
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