Balancing equations a) Cu_2O + C rightarrow Cu + CO b) Co(NO_3)_3 + (NH_4)_2S ri

Question

Balancing equations a) Cu_2O + C rightarrow Cu + CO b) Co(NO_3)_3 + (NH_4)_2S rightarrow Co_2S_3 + NH_4NO_3 c) (NH_4)_2CO_3 rightarrow CO_2 + H_2O + NH_3 d) N_2H_4 rightarrow NH_3 + N_2 e) FeS + HCl rightarrow FeCl_2 + H_2S Reaction Stoichiometry, Limiting Reactant, Theoretical and Percent Yields a) Use equation a from PART C-V. If 2.73 moles of C are reacted with an excess of Cu_2O, how many moles of CO would be produced? _______ moles b) Use equation b from PART C-V. If 79.2 g of Co(NO_3)_3 are reacted with an excess of (NH_4)_2S, how many moles of NH_4NO_3 would be produced? _______ moles c) Use equation from PART C-v. If 4.01 moles of (NH_4)_2CO_3 reacted, what is the theoretical yield of CO_2? _________ g d) Use equation d from PART C-V. If the theoretical yield is 28.0 grams of N_2, what mass in grams of N_2H_4, would have to react? ________ g e) Use equation e from PART C-V. If 35.0g of FeS is added to 35.0g of HCl and allowed to react, what is the theoretical yield of H_2S? _________ g Using the numbers from problem e above, if 10.2 g of H_2S were obtained in the lab (actual yield), what is the percent yield? ________ % g) How many grams of K_2C_2O_4, potassium oxalate, would contain 182 grams of pure potassium?

Explanation / Answer

a)

2.73 mol of C react to form CO:

Cu2O + C --> 2Cu + CO

ratio is 1:1 so

2.73 mol of C should produce 2.73 mol of CO

b)

2Co(NO3)3 + 3(NH4)2S --> Co2S3 + 6NH4NO3

79.2 g of Co(NO3)3

ratio is:

2 mol of Co(NO3)3 : 6NH4NO3

or 2:6 --> 1x3

MW of Co(NO3)3 = 244.9479

mol of Co(NO3)3 = mass/MW = 79.2 /244.9479 = 0.32333 g/mol

so

0.32333*3 = 0.96999 mol of NH4NO3 are expected

c)

from c:

(NH4)2CO3 --> CO2 + H2O + 2NH3

4.01 mol of reactant will produced 4.01 mol of CO2 due to 1:! ratio

mass of CO2 = mol*MW = 4.01*44 = 176.44 g

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