Question

1. Determine the volume (mL) required to prepare each of the following.

36.0 mL of a 0.500 M KNO 3 solution from a 5.50 M KNO 3 solution.

25.0 mL of 3.50 M H2SO4 solution using a 13.5 M H2SO4 solution.

0.360 L of a 2.50 M NH4Cl solution using a 12.0 M NH4Cl solution.

2.Answer the following for the reaction:
CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq)

How many milliliters of a 0.240 M HCl solution can react with 5.60 g of CaCO3 ?

How many liters of CO2 can form at STP when 17.5 mL of a 3.57 M HCl solution reacts with excess CaCO3 ?

What is the molarity of a HCl solution if the reaction of 210. mL of the HCl solution with excess CaCO3 produces 12.5 L of CO2 gas at 725 mmHg and 18∘C ?

Answer 1

1) a) C1 V1 = C2 V1

5.50M × V1 = 0.50 × 36ml

V1 = 0.50M × 36ml/5.50M

= 3.27ml

So, the answer is 3.27ml

b) 13.5M × V1 = 3.50M × 25ml

V1 = 6.48ml

So, the answer is 6.48ml

c)12.0M × V1 = 2.50M × 360ml

V1 = 2.50M × 360ml / 12.0ml

= 75ml

so, the answer is 75ml

2) CaCO3(s) + 2HCl(aq) -------> H2O(l) + CO2(g) + CaCl2(aq)

Molar mass of CaCO3 = 100.0869g

Mass of CaCO3 = 5.60g

No of Mole of CaCO3 = 5.60g/100.0869g = 0.05595

According to stoichiometry

1mole of CaCO3 react with 2 mole of HCl

so, 0.05595mol of CaCO3 require 0.1119mol of HCl

Volume of HCl required= (1000ml/0.240mol)×0.1119mol = 466.25ml

So, the answer is 466.25ml

b) No of mol of reacted = (3.57mol/1000ml)×17.5ml =0.062475ml

No of Mole of CO2 obtained= 0.062475mol

at STP 1mol gas occupy 22.4L

therefore ,

Volume occupied by 0.062475mol of CO2 = 22.4L × 0.062475= 1.3994L

So, the answer is 1.3994L

c) Ideal gas equation is

PV = nRT

no of mol ,n = PV/RT

No of mol of CO2 = 0.9539 atm × 12.5L / 0.082057(L atm /mol K ) ×291K

= 0.4993mol

0.4993mol of CO2 represents 0.9986mole of HCl

Therefore,

Molarity of HCl = (0.9986mol/210ml)×1000ml = 4.755M

so, the answer is 4.755M