Question

3. Predict the mass of Na2S2O3that could be produced from 38.2 g of Na2S and unlimited quantities of the other reactants, based on the following unbalanced equation:

Na2CO3+ Na2S + SO2--> Na2S2O3+ CO2

a.If the procedure produced only a 68.5 % reaction yield, how many grams of Na2S2O3would you expect to obtain?

Answer 1

We would obtain 79.4 grams of Na2S2O3

Step-by-step 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

Answer 2

1. 116.07g

2. 79.51g

Step-by-step 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%