Balancing equations CU_2O + C rightarrow 2 Cu + CO 2Co(NO_3)_3 + (NH_4)_2S right

Question

Balancing equations CU_2O + C rightarrow 2 Cu + CO 2Co(NO_3)_3 + (NH_4)_2S rightarrow Co_2S_4 + 6NH_4NO_3 (NH_4)_2CO_3 rightarrow CO_2 + H_2O + 2 NH_3 3N_2H_2 rightarrow 4 NH_3 + N_2 FeS + 2 HCl rightarrow FeCl_3 + H_2S Use equation a from PART 1. If 2.73 moles of C are reacted with an excess of Cu_2O, how many moles of CO would be produced? Use equation b from PART 1. If 79.2 g of Co (No_2)_2 are reacted with an excess of (NH_4)_2 S, how many moles of NH_2NO_3 would be produced? Use equation c from PART 1. If 4.01 moles of (NH_4)_2CO_3 reacted, what is the oretical yield of CO_2? Use equation d from PART 1. If the oretical yield is 28.0 grams of N_2, what mass in grams of N_2H_4, would have to react? Use equation e from PART 1. If 35. Og of FeS is added to 35.Og of HCl and allowed to react, what is the oretical yield of H_1S? Using the numbers from problem e above, if 10.2 g of H_2S were obtained m the lab (actual yield), what is the percent yield? How many grams of K_2C_2O_4, potassium oxalate, would contain 182 grams of pure potassium? Solution Concentration and Solution Stoichiometry What is the molarity of a solution made by dissolving 6.08g of sodium acetate, CH_3 COONa, in water, and diluting it to a total volume of 0.750L? M How many grams of strontium nitrate, Sr(NO_3)_2, are needed to prepare 3.40 Times 10^3 mL of a 0.925 M solution? g To what volume should 50.0mL of a 6.00M HCl stock solution be diluted to obtain a 1.50M HCl solution? How much water should be used? A 38.5mL sample of 0.875 M AgNO_2 solution is mixed with 51.0mL sample of 0.744M NaCl solution. The AgCl precipitated has a mass of 2.06 g. Calculate the limiting reactant, the oretical yield and percent yield.

Explanation / Answer

1)
a) Cu2O + C ----> 2 Cu + CO
b) 2 Co(NO3)3 + 3 (NH4)2S ------> Co2S3 + 6 NH4NO3
c) (NH4)2CO3 -----> CO2 + H2O + 2 NH3
d) 3 N2H4 ------> 4 NH3 + N2
e) FeS + 2 HCl -----> FeCl2 + H2S
2)
a) 1 mole of C produces 1 mole of CO then 2.73 moles of C produces 2.73 moles of CO
b) number of moles of Co(NO3)3 = 79.2/183 = 0.4328
2 moles of Co(NO3)2 produces 6 moles of NH4NO3
0.4328 moles of Co(NO3)2 produces 6/2*0.4328 = 1.2984
c) 1 mole of (NH4)2CO3 produces 1 mole of CO2
4.01 moles of (NH4)2CO3 produces 4.01 moles of CO2
d) 3 moles of N2H4 produces 1 mole of N2
3 * 32 g of N2H4 produces 28 g of N2
mass of N2H4 = 94 g of N2H4
e) number of moles of FeS = 35/88 = 0.3977 moles
number of moles of HCl = 35/36.5 = 0.9589 moles
1 mole of FeS produces 1 mole of H2S
0.3977 moles of FeS produces 0.3977 moles of H2S
2 moles of HCl produces 1 mole of H2S
0.9589 moles of HCl produces 0.9589/2 = 0.4794 moles
theoritical yield = 0.3977 moles
f) mass of H2S = 0.3977 * 34 = 13.5218 g
% yield = 10.2/13.5218*100 = 75.434 %
g) K2C2O4 contains 2 moles of K
number of moles of K = 182/39 = 4.67 moles
number of moles of K2C2O4 = 4.67/2 = 2.335 moles
mass of K2C2O4 = 2.335 * 166.22 = 388.1237 g

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