Respuesta :
Answer:
1. 116.07g
2. 79.51g
Explanation:
1. First, let us write a balanced equation for the reaction. This is illustrated below:
Na2CO3+ 2Na2S + 4SO2 —> 3Na2S2O3 + CO2
Next, let us calculate the mass of Na2S that reacted and the mass of Na2S2O3 produced from the balanced equation.
This is illustrated below:
Molar Mass of Na2S = (23x2) + 32 = 46 + 32 = 78g/mol
Mass of Na2S from the balanced equation = 2 x 78 = 156g
Molar Mass of Na2S2O3 = (23x2) + (32x2) + (16x3) = 46 + 64 + 48 = 158g/mol
Mass of Na2S2O3 from the balanced equation = 3 x 158 = 474g
From the balanced equation above,
156g of Na2S produced 474g of Na2S2O3.
Therefore, 38.2g of Na2S will produce = (38.2x474) /156 = 116.07g of Na2S2O3
From the calculations made above, 38.2g of Na2S will produce 116.07g of Na2S2O3.
2. The second part of the question shows that the percentage yield is 68.5%
Now, the grams we are expecting to have, talks about the experimental yield.
From the calculations made above, the theoretical yield is 116.07g. The experimental yield can be calculated for as shown below:
%yield = Experimental yield /Theoretical yield x100
68.5% = Experimental yield /116.07
68.5/100 = Experimental yield /116.07
Cross multiply to express in linear form as shown below:
100 x Experimental yield = 68.5x116.07
Divide both side by 100
Experimental yield = (68.5x116.07)/100
Experimental yield = 79.51g
Therefore, we are expected to obtain 79.51g of Na2S2O3 since the percentage yield is 68.5%
Answer:
We would obtain 79.4 grams of Na2S2O3
Explanation:
Step 1: Data given
Mass of Na2S = 38.2 grams
Molar mass Na2S = 78.05 g/mol
Molar mass Na2S2O3 = 158.11 g/mol
Reaction yield = 68.5 %
Step 2: The balanced equation
2Na2S + Na2CO3 + 4SO2 → 3Na2S2O3 + CO2
Step 3: Calculate moles Na2S
Moles Na2S = mass Na2S / molar mass Na2S
Moles Na2S = 38.2 grams / 78.05 g/mol
Moles Na2S = 0.489 moles
Step 4: Calculate moles Na2S2O3
For 2 moles Na2CO3 we need 1 mol Na2S and 4 moles SO2 to produce 3 mol Na2S2O3 and 1 mol CO2
For 0.489 moles Na2S we'll have 3/2 * 0.489 = 0.7335 moles Na2S2O3
If 0.7335 moles is produced the % yield is 100 %
For a 68.5 % yield, 0.7335 * 0.685 = 0.5024 moles
Step 5: Calculate mass Na2S2O3
Mass Na2S2O3 = 0.5024 moles * 15.11 g/mol
Mass Na2S2O3 = 79.4 grams
For a 100 % yield, we would obtain 115.97 grams of Na2S2O3
For a 68.5 % yield we would obtain 79.4 grams of Na2S2O3
