More practice from Chapters 11 and 12. 1. A piece of an unknown metal M weighing

Question

More practice from Chapters 11 and 12. 1. A piece of an unknown metal M weighing 1.42 g is dropped into aqueous HCl, producing MCl3 and H2 gas. 275 mL of H2 are collected at 22 °C and 745 mm Hg. What metal is M? Answer: I 2. Nitrogen gas and hydrogen gas react to make NH3 gas. How many L of NH3 gas at 45 ° C and 2.2 atm can be produced by reacting 125 L of nitrogen at at 25 °C and 4.0 atm with 275 L of hydrogen at at 25 °C and 5.0 atm? Answer: 440 L 3. How many mL of water do you have to add to 275 mL of 0.350 M NaCI to make a solution that is 0.100 M? Answer: 688 mL. 4. 0.254 g of an unknown acid are dissolved in 35.0 mL of water. It takes 18.4 ml. of 0.107 M NaOH to neutralize the acid. What is the molar mass of the acid? Answer 129 g/mol

Explanation / Answer

1) The reaction is

2M + 6HCl ------> 2MCl3 + 3H2(g)

3mol of H2 is produced from 2mole of Metal

So, if we know No of mole H2 produced we can calculate No of mole of Metal roduced

Ideal gas equation is

PV = nRT

Where,

P= Preaasure, 745mmHg=0.980atm

T= temperature, 22=295.15K

V = Volume , 275ml = 0.275L

R = gas constant , 0.082057(L atm/mol K)

no of mole , n = PV/RT

= 0.980atm × 0.275L/0.082057(L atm/mol K) × 295.15K

= 0.0111275mol

So, 0.0111275mol of H2 produced and it represents (2/3)×0.0111275 = 0.00742mol of Metal

Mass of Metal = 1.42g

Molar mass = mass/no of mol

= 1.42g/0.00742(g/ml)

= 191.37g/mol

this is nearest to molar mass(atomic mass) of Ir

Therefore,

The unknown metal is Ir

2) First determine the no of mole of Hydrogen and Nitrogen using ideal gas equation

No of mole,n=PV/RT

No of mole of Hydrogen =5.0atm × 275L/0.082057(L atm/mol K) × 298.15K = 56.20

No of mol of Nitrogen = 4.0atm × 125L/0.082057(L atm/mol K) × 298.15K =20.44

Now, look at the reaction

N2 + 3H2(g) -----> 2NH3(g)

Stoichiometrically , 1mole of N2 react with 3 mole of H2

So, 20.44mole of N2 react with 3× 20.44 = 61.32mole of H2 but avoilable H2 is 56.20only

So, H2 is limiting reagent

Stoichiometrically, 3mole of H2 produce 2mole of NH3, so 56.20mol of H2 produce (2/3)×56.20mol= 37.46moles of NH3

temperature of NH3 = 45=318.15K

Pressure of NH3 = 2.2atm

Volume, V = nRT/P

= 37.46mol×0.082057(L atm/mol K) ×318.15K/2.2atm

= 440L

3) Apply the formula

C1 × V1 = C2 × V2

where,

C1 = initial concentration , 0.350M

V1 = initial Volume , 275ml

C2 = Final concentration, 0.1M

V2 = Final volume,?

V2 = C1 × V1/C2

= 0.350M × 275ml/0.1M

= 962.5ml

Initial Volume, V1 = 275ml

Remaining water to be add = 962.5ml - 275ml = 688ml

4) The reaction is

HA + NaOH -----> NaA + H2O

1mole of NaOH react with 1mole of acid

No of mole of Consumed = (0.107mol/1000ml)×18.4ml = 0.00197mol

0.00197mol of NaOH react with 0.00197mol of Acid

Mass of acid = 0.254g

Molar mass = mass/no of mol

= 0.254g/0.00197mol

= 128.93g/mol

= 129g/mol

  

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.