The reaction of sulphuric acid with sodium hydroxide forming sodium sulphate and water (Neutralization reaction) ,

2NaOH(aq) + H2SO4(aq)------à Na2SO4(aq) + 2H2O (l)

1- Choose the correct answer

1- When 50 gm of CaCO3 decomposes thermally, ……..gm of CaO is formed (Ca = 40, C=12, O=16)

A- 28 B- 16 C- 76 D- 35

2- The volume of hydrogen required to form 11.2 L of water is……….

A- 22.4 L B- 11.2 L C- 68.2 L D- 44.8 L

3- One atomic unit equals …………. Gm

A- 1.66x10-24 B- 2.73x10-23 C- 1.75x10-15 D- 3.65x10-13

4- The unit used in IS for measuring the quantity of matter is…..

A- Mole B- Joule C- Calenda D- Kelvin

5- The mass of 44.8L of ammonia gas (NH3) in STP conditions is ………gm.

(N=14, H=1)

A- 0.5 B-2 C- 17 D- 34

6- If an amount of sodium has 3.01x1023 atoms, so its mass is ……..gm

A- 11.5 B- 0.5 C- 23 D-46

The chemical equation should be balanced according to..............

A- Avogadro's law B- Gay-Lussac's law C- law of mass conservation

D- Law of energy conservation

8- 0.5 mole of carbon dioxide gas (CO2) weighs….. gm (C=12, O=16)

A- 22 B- 44 C-66 D-88

9- When 64 gm of oxygen reacts with hydrogen, ….. litres of water vapour (H2O) are formed

A- 11.2 B- 22.4 C- 44.8 D- 89.6

10- The no. of moles in 36g of water equals…………. (H=1,O=16)

A- 1 B-2 C- 3 D- 4

11- The no. of molecules in 128g of sulphur dioxide (SO2) equals…… (S=32, O=16)

A- 2 B- 6.02x1023 C- 3.01x1023 D- 12.04x1023

12- The no. of sodium ions resulted from the dissolution of 40g of sodium hydroxide (NaOH) equals…..(Na=23,O=16,H=1)

A- 2 B- 6.02x1023 C- 3.01x1023 D- 12.04x1023

13- The volume of 4g of hydrogen in STP conditions equals …

A- 11.2 B- 22.4 C- 44.8 D- 89.6

14- The volumes of reactant gases are inversely proportional to those of products according to……….

A- Avogadro's law B- Gay-Lussac's law C- law of mass conservation

D- Law of energy conservation

Solve the following problems

1- Find the no. of sodium ions resulted from the dissolution of 117g of sodium chloride (NaCl) in water (Na=23,Cl=35.5)

2- 26.5g of sodium carbonate (Na2CO3) reacted with an abundant amount of hydrochloric acid in STP conditions (Na=23,C=12, O=16) find:-

a- The no. of water molecules

b- The volume of carbon dioxide gas

3- Calculate the no. of moles in 144gm. of carbon (C=12)

4- Calculate the mass of 2.4 moles of calcium carbonate (CaCO3) (Ca=40, C=12, O=16)

5- Calculate the volume of 56g of nitrogen gas in STP conditions (N=14)

6- 23g of sodium (Na) reacted with water(H2O) forming sodium hydroxide (NaOH) and hydrogen gas (Na=23, O=16, H=1), Find:-

a- The no. of sodium ions resulted from the reaction

b- The volume of evolving hydrogen gas

7- Calculate the molar mass of gaseous phosphorus in STP conditions, and the no. of atoms in one mole of it.

Balance the following equations

1- N2(g) + H2(g) àΔ NH3(g)

2- Cu(NO3)2(s) Δ à CuO(s) + NO2(g) + O2(g)

3- 3 Al(s) + O2(g) Δà Al2O3(s)

The Answers

Choose

1- 28gm

à The molar mass of calcium carbonate = 40 + 12 + 16+16+16 = 100 gm

The molar mass of calcium oxide (CaO) = 40+16 = 56

The no. of moles = the mass of the sample/ the molar mass = 50/100 = 0.5 mol

1 mole of calcium carbonate à 1 mole of calcium oxide CaO

0.5 mole of calcium carbonate à 0.5 mole of calcium oxide CaO

The mass of calcium oxide = 0.5 x molar mass = 0.5 x 56 = 28g

2- 11.2 L

2H2 + O2 à 2H2O

2 moles of hydrogen à 2 moles of water

The volume of hydrogen gas = the volume of water

The volume of water = 11.2 L

The volume of hydrogen gas = 11.2 L

3- 1.66x10-24

4- Mole

5- 34g

à The molar mass of ammonia gas = 14+1+1+1= 17gm

The no. of moles = the volume of gas in STP / 22.4 = 44.8/22.4 = 2 moles

The mass of 44.8 L of ammonia gas = no. moles x molar mass = 2 x 17 = 34g

6- 11.5gm

à the molar mass of sodium = 23 gm

The no. of moles = no. of atoms / Avogadro's number = 0.5 mol.

The mass of sodium = 0.5 x 23 = 11.5 gm

7- law of mass conservation

8- 22 g

à The molar mass of carbon dioxide gas = 16 + 16+12 = 44g

The mass of carbon dioxide = 0.5 x 44 = 22g

9- 89.6 L

2H2 + O2 à 2H2O

Molar mass of oxygen O2 = 16+16 = 32 gm

The no. of moles = 64 / 32 = 2 moles

1 mole of oxygen à 2 moles of water

2 moles of oxygen à 4 moles of water

The volume of water vapour in STP = 22.4 x the no. of moles = 22.4 x 4 = 89.6L

10- 2 moles

11- 12.04x1023 molecules

à The molar mass of sulphur oxide = 32 +16+16 = 64 g

The no. of moles = 128/64 = 2 moles

The no. of molecules = no. of moles x Avogadro's no. = 12.04x1023 molecules

12- 6.02x1023 ions

One mole of sodium hydroxide = 23 + 16+1 = 40g

One mole of NaOH à one mole of sodium ions

The no. of ions = 6.02x1023 ions

13- 44.8 L

à The molar mass of hydrogen gas = 2g

The no. of moles = 4/2 = 2 moles

The volume of gas in STP = 2 x 22.4 = 44.8 L

14- Gay-Lussac's law

Solve the following problems

1- The molar mass of sodium chloride = 23+35.5 = 58.5

The no. of moles in 117 gm of NaCl = 117/58.5 = 2 moles

One mole of NaCl à One mole of sodium positive ions

Two moles of NaCl à Two moles of sodium positive ions

The no. of sodium ions = 2 x 6.02x1023 = 12.04x10 23 ions

2- Na2CO3 +2 HCl à NaCl +H2O+ CO2

---

Дата добавления: 2015-10-23