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A mixture of pure BaCO3 and pure Na2CO3 weighs 1.000 g and has the total neutralizing power of 15.37 meq of CaCO3. Calculate the percentage of combined CO2 in the mixture and the weight of Li2CO3 that has the same neutralizing power as 1.000 g of the above mixture. Instructions: Include up to 4 decimal places See sample photo for the format of solution

A mixture of pure BaCO3 and pure Na2CO3 weighs 1.000 g and has the total neutralizing power of 15.37 meq of CaCO3. Calculate the percentage of combined CO2 in the mixture and the weight of Li2CO3 that has the same neutralizing power as 1.000 g of the above mixture. Instructions: Include up to 4 decimal places See sample photo for the format of solution

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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A mixture of pure BaCO3 and pure Na2CO3 weighs 1.000 g and has the total neutralizing power of 15.37 meq of CaCO3. Calculate the percentage of combined CO2 in the mixture and the weight of Li2CO3 that has the same neutralizing power as 1.000 g of the above mixture. Instructions: Include up to 4 decimal places See sample photo for the format of solution
Determination of the Proportion in which Components are present in a mixture: Example: If 0.5000 g of a mixture of CaCO3 and BaCO3 requires 30.00 mL of a 0.2500N HCI for neutralization. What is the percentage of each component? let x = g CaCO3 and y=g BaCO3 x + y = 0.5000 → equation1 eq of CaCO3 present = X CaCO3/2 50.04 eq of BaCO3 present = = BaCO3/2 eq CaCO3 + eq BaCO3 = eq HCI sample sol'n + 50.04 98.69 Solving equations 1 and 2 simultaneously: x = 0.247 g y = 0.253 g % CaCO3 = [0.247/0.500) x 100 = 49.4% % BaCO3 = (0.253/0.500) x 100 = 50.6% Example: The weight of combined LIOH, KOH and Ba(OH)2 in a mixture is 0.5000 g and requires 25.43 mL of 0.500N acid for neutralization. The same weight of sample with CO₂ gives a precipitate of BaCO3 that requires 5.27 mL of the above acid for neutralization. Find the weights of LiOH, KOH and Ba(OH)2 in the original mixture. Let x = g LIOH and y = g KOH and z= g Ba(OH)2 X + y + z = 0.5000 → 1 eq LiOH = X eq Ba(OH)2 = Z 23.95/1 171.36/2 eq KOHV 56.11/1 eq LIOH + eq KOH + eq Ba(OH)2 = 25.43mL (0.5eq/L) 1000mL/L → equation 2 eq Ba(OH)2 = eq BaCO³ = 5.27 ml (0.5 eq/l) 1000mL/L wt Ba(OH)2 x 2 eq/mol = 2.635 x 10³ eq Sample Soln wt Ba(OH)2 = 0.2258 g X x + y = 0.2742 eq LIOH + eq KOH + eq Ba(OH)2 = 25.43mL[0.5eq/L) 1000mL/L → equation 2 eq LIOH + eq KOH +2.635 x 103 eq = 0.012715 eq X + y = 0.01008 23.95/1 56.11/1 Solving equations 1 and 2 simultaneously: x = 0.217 g y = 0.0572 g Question: A mixture of pure BaCO3 and pure Na₂CO3 weighs 1.000 g and has the total neutralizing power of 15.37 meq of CaCO3. Calculate the percentage of combined CO₂ in the mixture and the weight of Li₂CO3 that has the same neutralizing power as 1.000 g of the above mixture. 98.69 = 30.0 mL (0.250eq/L) (1000mL/L) → equation2 171.36 g/mol + y + 0.2258 0.5000 →1
Transcribed Image Text:Determination of the Proportion in which Components are present in a mixture: Example: If 0.5000 g of a mixture of CaCO3 and BaCO3 requires 30.00 mL of a 0.2500N HCI for neutralization. What is the percentage of each component? let x = g CaCO3 and y=g BaCO3 x + y = 0.5000 → equation1 eq of CaCO3 present = X CaCO3/2 50.04 eq of BaCO3 present = = BaCO3/2 eq CaCO3 + eq BaCO3 = eq HCI sample sol'n + 50.04 98.69 Solving equations 1 and 2 simultaneously: x = 0.247 g y = 0.253 g % CaCO3 = [0.247/0.500) x 100 = 49.4% % BaCO3 = (0.253/0.500) x 100 = 50.6% Example: The weight of combined LIOH, KOH and Ba(OH)2 in a mixture is 0.5000 g and requires 25.43 mL of 0.500N acid for neutralization. The same weight of sample with CO₂ gives a precipitate of BaCO3 that requires 5.27 mL of the above acid for neutralization. Find the weights of LiOH, KOH and Ba(OH)2 in the original mixture. Let x = g LIOH and y = g KOH and z= g Ba(OH)2 X + y + z = 0.5000 → 1 eq LiOH = X eq Ba(OH)2 = Z 23.95/1 171.36/2 eq KOHV 56.11/1 eq LIOH + eq KOH + eq Ba(OH)2 = 25.43mL (0.5eq/L) 1000mL/L → equation 2 eq Ba(OH)2 = eq BaCO³ = 5.27 ml (0.5 eq/l) 1000mL/L wt Ba(OH)2 x 2 eq/mol = 2.635 x 10³ eq Sample Soln wt Ba(OH)2 = 0.2258 g X x + y = 0.2742 eq LIOH + eq KOH + eq Ba(OH)2 = 25.43mL[0.5eq/L) 1000mL/L → equation 2 eq LIOH + eq KOH +2.635 x 103 eq = 0.012715 eq X + y = 0.01008 23.95/1 56.11/1 Solving equations 1 and 2 simultaneously: x = 0.217 g y = 0.0572 g Question: A mixture of pure BaCO3 and pure Na₂CO3 weighs 1.000 g and has the total neutralizing power of 15.37 meq of CaCO3. Calculate the percentage of combined CO₂ in the mixture and the weight of Li₂CO3 that has the same neutralizing power as 1.000 g of the above mixture. 98.69 = 30.0 mL (0.250eq/L) (1000mL/L) → equation2 171.36 g/mol + y + 0.2258 0.5000 →1
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