Question

1. A 1.500-g sample of potassium hydrogen carbonate is decomposed by heating to produce 1.040 g of potassium carbonate. Calculate the theoretical yield and percent yield of K CO. 2 KHCO3() 4 K2CO3(s) + H2O(g) + CO2(g) 1.500, KHEO3(-)x mo Brux Imot tos 138.2.109 K2CO3 100.129 Hle, 2 mot Kaso, x mol becos 1.0351x100 -0.996 2 (100% 1.040g 1.035 g K2CO3 - 100% 2. A 1.750-g sample containing potassium hydrogen carbonate is decomposed by heating. If the mass loss is 0.271 g. what is the percentage of KHCO, in the unknown mixture? 2 KHCO3(s) 4 K2CO3(s) + H2CO3(g) Decomposing Baking Soda
Is this answer correct? Also, please help me with the 2nd question.

Answer 1

Question 1

Balanced equation:
2 KHCO₃ ====> K₂CO₃ + H₂O + CO₂
Reaction type: decomposition

Mass of KHCO3 = 1.5 gm

Moles of KHCO3 = 1.5 gm / 100.115 gm/mol = 0.01498 Moles

Moles of K2CO3 produced =  0.0074913 Moles

Mass of K2CO3 produced =  0.0074913 Moles x 138.2055 g/mol = 1.0353 gm

The actual yield is 1.040 gm. Hence Theoretical yield is 100 %

Question 2

Balanced equation:
2 KHCO₃ ===> K₂CO₃ + H₂CO₃
Reaction type: decomposition

The loss of mass is 0.271 gm which is H2CO3.

Mass of H2CO3 = 0.271 gm

Moles of H2CO3 = 0.271g/ 62.02 g/mmol =  0.004369 Moles

Moles of KHCO3 reacted =  0.00873 Moles

Mass of KHCO3 presents = 0.00873 Moles x 100.115 g/mol = 0.8748 gm

Percentage of KHCO3 in the unknown mixture = 0.8748 gm x100% /1.750 gm = 49.988 %

49.988 % of KHCO3 is presents in unknown mixture.

Balanced equation: 2 КНСОЗ 3 К2СО3+ H-СОз Reaction type: decomposition Reaction stoichiometry Limiting reagent Compound Coefficient Molar Mass Moles Weight 0.008738442 0.8748503 КНСО 2 100.11514 K2CO3 .00436922 0.6038503 1 138.2055 H2CO3 0.2710000 0.004369221 1 62.02478 Units: molar mass - g/mol, weight - g.