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stoichiometry exercises:

1) Analysis of a copper sulfide (CuS) indicates that 65.8% by weight of the sulfide corresponds to the metal. Calculate the amount in grams of copper sulfide that can be obtained from 42.0 grams (g) of sulfur (S).




2) When analyzing an oxide of copper, it was found to contain 79.9% by weight of copper. Calculate the amount in grams of oxide that will be formed from:


a) 25.0 g of oxygen

b) 25.0 g of copper

c) If you have 25.0 g of copper and 25.0 g of oxygen



3)When analyzing a compound formed by elements A and B, they have been found to combine in an A / B mass ratio of 2.00 / 7.00 respectively. Calculate:


a) How many grams of B are needed for 25.0 grams of compound to form?

b) How many grams of A are needed for 25.0 grams of compound to form?

c) How many grams of A are required to fully react 0.500 grams of B?

d) If you have 80.0 grams of A and 80.0 grams of B, how many grams of product are obtained?

e) What is the percentage composition by mass of the compound?



4)A stone the size of a tennis ball weighs approximately 1.05 kg. How many moles of stone will it take to equal the mass of planet Earth, which is approximately 6 x 1027 g?



5)Ammonium nitrate, NH4NO3, which is prepared from nitric acid, is used as a nitrogen fertilizer. Calculate the mass percent of the elements in ammonium nitrate?



6)1.285 g of an organic compound containing carbon, hydrogen and oxygen are burned in a laboratory. Combustion resulted in 1.45 g of CO2 and 0.890 g of H2O. What is the empirical formula? (Data: atomic mass C = 12.0 g mol-1; O = 16.0 g mol-1; H = 1.00 g mol-1)



7)Under certain experimental conditions, chlorine, Cl2, reacts with oxygen, O2, forming chlorine oxide (V), Cl2O5. In this regard, write the equation that represents the chemical reaction and solve: (Data: atomic mass Cl = 35.5 g mol-1


a) If we have 50.0 g of chlorine and 50.0 g of oxygen, how many g of Cl2O5 are formed?

b) What element is in excess?

c) How many g of the excess element remain unreacted?



8) The fermentation of sugar generates ethyl alcohol and carbon (IV) oxide according to the following unbalanced equation.


C6H12O6   →     C2H5OH  +  CO2



Balance the equation and determine the mass of CO2 that is formed when 9.53 g of C2H5OH is produced in the fermentation of sugar.




9)the phosphorus, P4, burns with excess oxygen, O2, to form phosphorous oxide (V), P2O5. In this regard: (Data: atomic mass P = 31.0 g mol-1 )


a) balance the chemical equation for the given combustion.

b) How many moles of P2O5 are formed due to the combustion of 6.00 g of P4?

c) P2O5 in contact with excess water reacts to form H3PO4. Write the corresponding chemical equation.



10) A mass of 1.84 g of NaOH is dissolved in water and 2.05 g of gaseous H2S are bubbled in, the following reaction occurring: (Data: PA Na = 23.0 g mol-1 S = 32.0 g mol-1)


H2S  +  2NaOH        →        Na2S  +  2H2O


What mass of Na2S is produced, assuming the limiting reagent is completely consumed?




11)A student reacts benzene, C6H6, with bromine, Br2, in an attempt to prepare bromobenzene, C6H5Br, according to the reaction:


C6H6  +  Br2           →                C6H5Br  +  HBr (Data: atomic mass Br = 80,0 g mol-1)



a) What is the theoretical yield of bromobenzene in this reaction if 30.0 g of benzene react with an excess of Br2?

b) What is the percentage yield if 56.7 g of bromobenzene are obtained?



12) In aqueous solution, ferric chloride reacts with an excess of ammonia and water, according to the following equation:


FeCl3  +  NH3 + H2O     →       Fe(OH)3  +  NH4Cl (Data: atomic mass Fe = 56,0 g mol-1 N = 14,0 g mol-1)


In this regard, balance the equation and answer the following questions:


a) What is the amount in grams of Fe (OH) 3 that is formed by reacting 20.0 g of ferric chloride?

b) Calculate the number of ammonium chloride molecules formed from 20.0 g of ferric chloride.

c) How many moles of ammonia are needed for the 20.0 g of ferric chloride to fully react?



13) Chlorine gas can be prepared from the following reaction:


MnO2   +   4 HCl       →   Cl2  +  MnCl2   +  2 H2O


Assuming that the limiting reagent of the reaction is MnO2. The grams of Cl2 that can be produced from 25.0 g of MnO2 are: 


Limiting reagent MnO2: a) 0.0326 b) 10.2 c) 20.4 d) 30.6 e) 246



14) From 0.75 mol of MnO2 and 2.0 mol of HCl, in number of moles of Cl2 that    can be prepared are:


a)   2,0

b)   1,3

c)   1,0

d)   0,75

e)   0,50

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