National Chemistry

Calculations from equations. THE QUESTIONS (answers below)

2000
18. Calculate the mass of water produce when 22g of propane are completely burned in air.

C3H8 + 5O2 -----> 3CO2 + 4H2O
 
2001
12. Calculate the mass of sulphur produced, in grams, when 34g of hydrogen sulphide reacts with sulphur dioxide.
2H2S + SO2 -------> 3S + 2H2O

2002
Q15 Calculate the mass of iron, in tonnes, which is produced from 160 tonnes of iron(III) oxide.

  Fe2O3 + 3CO -----> 2Fe + 3CO2

2003
12. Calculate the mass of zinc required to produce 0.5 mole of hydrogen.

Zn + 2HCl ----> ZnCl2 + H2
2004
nil

2005
20. Calculate the mass of hydrogen produced when 1.35g of aluminium reacts with sulphuric acid.

2Al + 3H2SO4 -----> Al2(SO4)3 + 3H2

2006
15. Calculate the mass of oxygen produced when 22g of dinitrogen monoxide decomposes

2N2O ---------> 2N2 + O2

2007
20. Calculate the mass of calcium carbonate required to neutralise 0.2 mole of hydrochloric acid

CaCO3 + 2HCl -----> CaCl2 + H2O + CO2

2008
20. Calculate the mass of iron produced from 40g of iron oxide
2Al + Fe2O3 ----> 2Fe + Al2O3

2009
19. Calculate the mass of water produced when 6.4g of nonane is burned.

C9H20 + 14O2 -----> 9CO2 + 10H2O

2010
12. Calculate the mass of hydrogen produced when 4.9g of magnesium reacts with an excess of dilute sulphuric acid.

Mg + H2SO4 ____> MgSO4 + H2

 

  answers

2000

step 1: C3H8 + 5O2 -----> 3CO2 + 4H2O
step 2: 1 mole -----> 4 mole
step 3
  mass 1 mole C3H8 = (3 x C) + (8 x H)
  (3 x 12) + (8 x 1) = 44g

  mass 1 mole H2O = (2 x H) + (1 x O)
  = (2 x 1) + (1 x 16)= 18g

  C3H8  + 5O2 -----> 3CO2 + 4H2O
  1 mole -----> 4 mole
  44g 4 x 18g
  = 72g
 
  So 44g of propane makes 72 g of water

therefore 
  1g produces more than 1g but not as much as 72g

calculate  44 ----> 72
1 ----> 72  (=1.63)
  44

  22 ---> 72 x 22  = 36g
  44

2001
step 1: 2H2S + SO2 -------> 3S + 2H2O
step 2: 2 mole of H2S -------> 3 mole of S
step 3a (2 x H)+(1 x S) -------> (3 x S)
(2 x 1)+(1 x 32) -------> (3 x 32)
step3b: 2 mole x 34g ------->3 mole x 32
68g -------> 96g
so 68g of H2S makes 96g of S

therefore  1g of H2S makes more than 1g of S

calculate  68g -------> 96
  1g -----> 96  (1.41)
  68

  34g -----> 96  x 34 = 48g
  68

2002
step 1: Fe2O3  + 3CO -----> 2Fe  + 3CO2
step 2: 1 mole of Fe2O3 -----> 2 moles of Fe
step 3a:
  (2 x Fe)+(3 x O) -----> (2 x Fe)
step 3b:
  (2 x 56)+(3 x 16) -----> (2 x 56)
160 -----> 112
  thus 160g iron oxide makes 112g iron
step 4: units are not g!
  but 160tonnes iron oxide makes 112 tonnes iron
 
no calculation needed !!!

 
2003
12. Calculate the mass of zinc required to produce 0.5 mole of hydrogen.

step 1: Zn + 2HCl ----> ZnCl2 + H2
step 2: 1 mole 1 mole
step 3a: (1 x Zn) not in grams
  (1 x 65.5) = 65.5 1 mole
3b: not needed
  65.5g -----> 1 mole
therefore  65.5g Zn produces 1mole H2

  1g Zn would produce less than 1mole of H2

calculate  65.5g <--- 1 mole
  32.75 <--- 0.5 mole

2005
step 1: 2Al + 3H2SO4 -----> Al2(SO4)3 + 3H2
step 2: 2 moles of Al -----> 3 moles of H2
step 3a: (1 x Al) -----> (2 x H)
  ( 1 x 27)=27g (2 x 1)=2g
step3b: 2 x 27g ---> 3 x 2g
  54g ---> 6g

therefore 54g Al produces 6g H2

  1g Al would produce less than 1g of H2

calcualte 54g ---> 6g
  1g ---> 6  = 0.11
  54

1.35 ---> 6  x 1.35 =0.15g
  54
 
 

2006
1: 2N2O  ---------> 2N2 + O2
2: 2 moles 1 mole
3a: (2 x N)+(1 x O) (2 x O)
(2 x 14)+(1 x 16) (2 x 16)
  = 44g = 32g
3b: 2 x 44g 32g
  88g ----> 32g

thus 88g of N2O produces 32g of O2

therefore 1g of N2O produces less than 1g of O2

calculate 88g ----> 32g
  1g ----> 32  (0.36)
  88

22g ----> 32  x 22 = 7.99
  88

2007

1: CaCO3  + 2HCl  -----> CaCl2 + H2O + CO2
2: 1 mole 2 mole

beware (i) working backwards and HCl not in grams

3a: CaCO3 = (1 x Ca)+(1 x C)+(3 x O)
(1 x 40)+(1 x 12)+(3 x 16) = 100g

HCl is not in grams so can leave as 0.2 mole

thus 1 mole of CaCO3 reacts with 2 mole of HCl
  ie 100g of CaCO3 reacts with 2 mole of HCl

calculate 100g <------ 2 mole

  100  <------ 1 mole
  2
 
100 x 0.5 <------ 0.5 mole
  2

  = 25g

2008
Calculate the mass of iron produced from 40g of iron oxide
1: 2Al +  Fe2O3  --->  2Fe  + Al2O3
2: 1 mole 2 mole
3a: (2 x Fe)+(3 x O) (1 x Fe)
  (2 x 56)+(3 x 16) (1 x 56)
  = 160g = 56
3b: 160g 2 x 56
  160g 112g

thus 160g Fe2O3 produces 112g Fe

therefore 1g Fe2O3 produces les than 1g

calculation 160 -------> 112
  1 -------> 112  (0.7)
  160

  40 -------> 112  x 40 = 28g
  160

2009
Calculate the mass of water produced when 6.4g of nonane is burned.
step 1: C9H20  + 14O2 -----> 9CO2 + 10 H2O
step 2: 1 mole -----> 10 mole
step 3
mass 1 mole C9H20 = (9 x C) + (20 x H)
  (9 x 12) + (20 x 1) = 128g

mass 1 mole H2O = (2 x H) + (1 x O)
  = (2 x 1) + (1 x 16)= 18g

   C9H20 + 14O2 -----> 9CO2 +  10 H2O
  1 mole -----> 10 mole
  128g 10 x 18g
  = 180g
  So 128g of nonane makes 180 g of water
therefore  1g produces more than 1g
calculate  128 ----> 180
1 ----> 180  (=1.4)
  128

  6.4 ---> 180  x 6.4 = 8.96g
  128

2010
Calculate the mass of hydrogen produced when 4.9g of magnesium reacts with an excess of dilute sulphuric acid.

1: Mg  + H2SO4 ----> MgSO4 +   H2
2: 1 mole 1 mole
3a: (1 x Mg) (2 x H)
  (1 x 24) (2 x 1)
  = 24g = 2g
3b: (not required)
 
thus   24g of Mg produces 2g of H2

therefore  1 g of Mg produces less than 1g of H2

calculate  24 -----> 2
  1 ----->  2   (0.083)
  24

  4.9 -----> 2  x 4.9 = 0.4g
  24

   

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