Question

1. How many mL of 0.644 M HBr are needed to dissolve 6.76 g of CaCO3? 2.Calculate the number of milliliters of 0.505 M Ba(OH)2 required to precipitate all of the Cu2+ ions in 179 mL of 0.656 M CuCl2 solution as Cu(OH)2. The equation for the reaction is: 3. What volume of a 0.334 M hydrobromic acid solution is required to neutralize 15.7 mL of a 0.183 M barium hydroxide solution? 4. What volume of a 0.161 M calcium hydroxide solution is required to neutralize 19.1 mL of a 0.205 M hydrochloric acid solution? 5. What volume of a 0.292 M perchloric acid solution is required to neutralize 28.9 mL of a 0.126 M potassium hydroxide solution?

Answer 1

1. 2Hbr + CaCO3 => CaBr2 + CO2 + H2O

1 mol CaCO3 = 40+12+3*16 = 100g

So, 6.67g CaCO3 = 6.67/100 mol

1 mol CaCO3 requires 2 moles of HBr

So, 6.67/100 mol of CaCO3 requires 0.1334 mol of HBr

Now, 1000 ml contains .644 moles of HBr

So, 1 mol of HBr is in 1000/0.644 ml of the solution

So, 0.1334 mole of HBr is in 1000/0.644*0.1334 ml of Sol = 207.14 ml

2. CuCl2 + Ba(OH)2 => Cu(OH)2 + BaCl2

Number of moles of CuCl2 = 0.179L*0.656M = 0.117424 moles

So, required Ba(OH)2 is 0.117424 moles

Hence, the volume of 0.505M Ba(OH)2 required = 0.117424/0.505 = 232.5 ml

3. 2HBr + Ba(OH)2 => BaBr2 + 2H2O

15.7 ml of 0.183 M Ba(OH)2 contains 0.0157*0.183 moles of Ba(OH)2 = 2.873*10^-3 moles

So, number of moles of HBr required for complete reaction is 2*2.873*10^-3 moles = 5.7462*10^-3 moles

Hence, volume of 0.334 M HBr required = 5.7462*10^-3/0.334*1000 ml = 17.2 ml

4. HCl + Ca(OH)2 => CaCl2 + H2O

19.1 ml of 0.205 M HCl contains 0.0191*0.205 moles = 3.9155*10^-3 moles

So, number of moles of Ca(OH)2 required for complete reaction is 3.9155*10^-3 moles

Hence, volume of 0.161 M Ca(OH)2 required = 3.9155*10^-3/0.161*1000 ml = 24.32 ml

5. HCLO4 + KOH => KClO4 + H2O

28.9 ml of 0.125 M KOH contains 0.0289*0.125 moles = 3.6125*10^-3 moles

So, number of moles of HCLO4 required for complete reaction is 3.6125*10^-3 moles

Hence, volume of 0.292 M HCLO4 required = 3.6125*10^-3/0.292*1000 ml = 12.37 ml