Could you go step by step of how did you get the answers please. Thank you! The

Question

Could you go step by step of how did you get the answers please. Thank you!

The reaction in question #15 is carried out to completion in the laboratory and produces 1.92 moles of H_2O. What is the percent yield of the reaction? A) 77% B) 96% C) 11% D) 43% E) Impossible to determine with the given information. Consider the following balanced reaction: C_3H_8(g) + 5 O_2(g) rightarrow 3 CO_2(g) + 4 H_2O(g) If 2.5 moles of O_2 is reacted with 4.6 moles of C_3H_8, what is the theoretical yield of H_2O in grams? What is the limiting reactant? A) 36.0 g H_2O: O_2 is limiting reactant B) 36.0 g H_2O: C_3H_8 is limiting reactant C) 332 g H_2O: O_2 is limiting reactant D) 332 g H_2O: C_3H_8 is limiting reactant E) 2.5 g H_2O: O_2 is limiting reactant Write a balanced net ionic equation for the reaction that may occur when AgNO_3(aq) is mixed with Na_2CO_3(aq). A) No reaction B) 2 Na^+(aq) + 2 NO_3^-(aq) rightarrow 2 NaNO_3(s) C) 2 Ag^+(aq) + CO_3^2-(aq) rightarrow Ag_2CO_3(s) D) 2 Na^+(aq) + 2 Ag^+(aq) + 2 NO^3-(aq) + CO_3^2-(aq) rightarrow Ag_2CO_3(s) + 2 NaNO_3(s) E) Ag^+(aq) + NO_3^- (aq) rightarrow AgNO_3(s) The combustion of methane (CH_4) is: CH_4 (g) + 2O_2(g) rightarrow 2H_2O(l) + CO_2(g) How many grams of water are produced upon the combustion of 12.3 grams of methane? A) 36.0 g B) 27.6 g C) 13.8 g D) 8.0 g E) 24.6 g

Explanation / Answer

C3H8 + 5O2 --------------> 3CO2 + 4H2O

1 mole of C3H8 react with 5 moles of O2

4.6 moles of C3H8 react with = 5*4.6/1   = 23 moles of O2 is limiting reagent

5 moles of O2 react with 1 moles of C3H8

2.5 moles of O2 react with = 1*2.5/5   = 0.5 moles of C3H8

   O2 is limiting reagent

C3H8 + 5O2 --------------> 3CO2 + 4H2O

5 moles of O2 react with C3H8 to gives 4 moles of H2O

2.5 moles of O2 react with C3H8 to gives = 4*2.5/5   = 2 moles of H2O

Theoritical yield of H2O = 2*18 = 36g >>>>>answer ......A

16

Actual yield of yiled of H2O = 1.92*18   = 34.56g

percentage yiled = actual yield*100/theoritical yield

                             = 34.56*100/36   = 96% >>>>>>B

20. Na2CO3 (aq) + 2AgNO3 (aq) --------------> Ag2CO3(s) + 2NaNO3(aq)

   2 Na^+(aq)+CO3^- (aq) + 2Ag^+(aq) +2NO3^- (aq) --------------> Ag2CO3(s) + 2Na^+(aq) +2NO3(aq)

removal of spectator ions to get net ionic equation

CO3^- (aq) + 2Ag^+(aq) --------------> Ag2CO3(s)   net ionic equation >>>answer ....>C

17 . CH4 + 2O2 ----------> CO2 + 2H2O

    1 mole of CH4 combustion to gives 2 moles of H2O

   16g of CH4 combustion to gives 2*18g of H2O

12.3g of CH4 combustion to gives = 2*18*12.3/16   = 27.675g of H2O >>>>>answer ........B

   

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