Question

In a solution of Nicly, the cl' concentration is 0.017 M. How many grams of Nicly are in 538 mL of this solution? g NICI the tolerance is +/-2% For the reaction: Na2CO3(aq) + NiCl2(aq) → NICO3(s) + 2NaCl(aq) How many milliliters of 0.38 M NiCl solution are needed to react completely with 31.4 mL of 0.40 M Na2CO3 solution? mL NICI How many grams of Nicoz will be formed? g NICO: What is the molarity of an aqueous solution of potassium hydroxide if 25.12 mL is exactly neutralized by 24.05 mL of 0.148 M HCI? M How many milliliters of 0.297 M FeCl3 solution are needed to react completely with 24.2 mL of 0.0224 M AgNO3 solution? The net ionic equation for the reaction is: Ag+(aq) + Cl(aq) + AgCl(s) mL of FeCl3: ml FeCl3 How many grams of AgCl will be formed? g AgCI

Answer 1

Given, concentration of Cl" = 0.017 M volume of the solution = 538 mL Calcualte the moles of Cl as follows: moles of C1° = concentration of Clºx volume of the solution R0017 1 mol/L 520 - 1L 1000 mL = 9.1x10 mol ----------------- -------------------- Ionisation of NiCl, is as follows: Nici —— Ni2+ + 2C1- Stoichiometric relation between NiCl, and C1 is moles of NICI. = moles of a- 9.1x10' mol moles of NaCl, = 2 = 4.55x10-mol molar mass of NiCl, = 129.6 g/mol mass of NiCl, = moles of NiCl, Xmolar mass of Na, = 4.55x10-3 mol x129.6 g/mol = 590x10 g = 0.59 g Therefore, the required solution is 10.59g|Na, Write the reaction as follows: Na,coz (aq) + Nici, (aq) →→NCO (s) + 2NaCl(aq) concentration of Na,Co, = 0.40 M volume of Na,C0, = 31.4 mL concentration of NiCl, = 0.38 M volume of NiCl = ?

From the reaction, moles of Na,Co, = moles of Nici, concentration of Na,Co, volume of Na,co; = concentration of Nia, xvolume of NICI, 0.40 MX31.4 mL = 0.38 Mxvolume of Nici, 0.40 Mx31.4 mL volume of NiCl, = ? 0.38 M = 33 mL Therefore, the required solution is 33 mL --- -- -- -- -- -- -- -- -- -- -- -- -- -- --- moles of NiCO, formed = moles of Na,co = 0.40 Mxmol/L 11 1000 mL M *31.4 ml = 0.013 mol molar mass of NiCO, = 118.7 g/mol mass of NiCO, = 0.013 mol x118.7 g/mol = 1.5g Therefore, the required solution is 1.5g|NCO3. Write the reaction potassium hydroxide and HCl as follows: KOH(aq) +HCl(aq) →→ Ka(s) + H20(1) concentration of KOH = ? volume of KOH = 25.12 mL concentration of HCl = 0.148 M volume of HCl = 24.05 mL for neutralization, moles of KOH= moles of HCI concentration of KOH Xvolume of KOH = concentration of HC1 x volume of HC1 concentration of KOH X 25.12 mL = 0.148 MX 24.05 mL 0.148 Mx24.05 mL concentration of KOH= 25.12 mL = 0.142 M Therefore, the required solution is 0.142 M