The Solvay process for the manufacture of sodium carbonate begins by passing ammonia and carbon dioxide through a solution of sodium chloride to make sodium bicarbonate and ammonium chloride. The equation for this reaction is H2O + NaCl + NH3 + CO2 → NH4Cl + NaHCO3. In the next step, sodium bicarbonate is heated to give sodium carbonate and two gases, carbon dioxide and steam 2NaHCO3 → Na2CO3 + CO2 + H2O
What is the theoretical yield of sodium carbonate, expressed in grams, if 144 g of NaCl were used in the first reaction?
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If 87.6 g of Na2CO3 were obtained from the reaction described in part a, what was the percentage yield?
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I got answer 130.57 g of theoratical yield
and 67% of percentage yield.
Pleas can u check me , my answer is correct or not. Thanks
130.57 g for the theoretical yield looks ok but you aren't allowed that many significant figures. You are allowed three (from the 144); therefore, I would round the theoretical yield to 131 g. For percent yield, You are allowed three and you have only two; therefore, I would recalculate that part of it and round to three places. I have 66.9%.
Determine the mass of antimony produced when 0.46g of antimony (III) oxide reacts with carbon.... (cont.)?
....occording to the following equation
Sb2O3+3C ---> 2Sb+3CO
please help me im soo lost, thank you!
To find the theoretical yield of sodium carbonate, you'll need to follow these steps:
Step 1: Determine the molar mass of NaCl (sodium chloride). The molar mass of Na is 22.99 g/mol, and the molar mass of Cl is 35.45 g/mol. Adding them together, we get:
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Step 2: Convert the mass of NaCl to moles by dividing the given mass by the molar mass:
Moles of NaCl = 144 g NaCl / 58.44 g/mol = 2.47 mol NaCl
Step 3: Since the reaction equation shows a 1:1 molar ratio between NaCl and Na2CO3, the moles of NaCl also represent the moles of Na2CO3 formed. So the theoretical yield of Na2CO3 is 2.47 mol.
Step 4: Determine the molar mass of Na2CO3 (sodium carbonate). The molar mass of Na is 22.99 g/mol, the molar mass of C is 12.01 g/mol, and the molar mass of O is 16.00 g/mol. Adding them together, we get:
Molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol
Step 5: Convert the moles of Na2CO3 to grams by multiplying the moles by the molar mass:
Theoretical yield of Na2CO3 = 2.47 mol × 105.99 g/mol = 260.92 g
So the theoretical yield of sodium carbonate, expressed in grams, is 260.92 g (not 130.57 g as you mentioned).
To determine the percentage yield, you'll need to know the actual yield of Na2CO3, which is given as 87.6 g.
Percentage yield = (Actual yield / Theoretical yield) × 100
Percentage yield = (87.6 g / 260.92 g) × 100 = 33.59%
So the correct percentage yield is approximately 33.59% (not 67% as you mentioned).
Please double-check your calculations.