Answer:17.65g
Explanation:
Use ratio and calculate from the balanced equation
For the reaction between carbon disulfide and oxygen to produce carbon dioxide and sulfur dioxide, we have:
1. The balanced chemical reaction is:
CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g)
2. The molar mass of all four chemicals is:
[tex] M_{CS_{2}} = 76.131 g/mol [/tex]
[tex] M_{O_{2}} = 31.998 g/mol [/tex]
[tex] M_{CO_{2}} = 44.009 g/mol [/tex]
[tex] M_{SO_{2}} = 64.058 g/mol [/tex]
3. The number of grams of sulfur dioxide produced when a mixture of 10.5 g of carbon disulfide and 41.6 g of oxygen reacts is 17.68 grams.
1. The chemical equation for the reaction between carbon disulfide and oxygen gas to produce carbon dioxide and sulfur dioxide is the following:
CS₂(g) + O₂(g) → CO₂(g) + SO₂(g) (1)
To balance the reaction we need to equal the number of C, S, and O atoms in the reactants and products. The number of C atoms is balanced, but in the reactants, we have 2 S atoms and on the products, we have 1, so we need to add a coefficient of 2 before the SO₂ (products side).
CS₂(g) + O₂(g) → CO₂(g) + 2SO₂(g) (2)
Now, we can see that we have 6 O atoms on the products side and 2 O atoms on the reactants, so we need to add a coefficient of 3 before the O₂ (reactants side).
CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g) (3)
The reaction (3) is balanced.
2. The molar mass of the 4 chemicals can be calculated as follows:
[tex]M_{CS_{2}} = A_{C} + 2A_{S} = 12.011 g/mol + 2*32.06 g/mol = 76.131 g/mol[/tex]
[tex]M_{O_{2}} = 2A_{O} = 2*15.999 g/mol = 31.998 g/mol[/tex]
[tex]M_{CO_{2}} = A_{C} + 2A_{O} = 12.011 g/mol + 2*15.999 g/mol = 44.009 g/mol[/tex]
[tex]M_{SO_{2}} = A_{S} + 2A_{O} = 32.06 g/mol + 2*15.999 g/mol = 64.058 g/mol[/tex]
Hence, the molar mass of CS₂, O₂, CO₂, and SO₂ is 76.131 g/mol, 31.998 g/mol, 44.009 g/mol, and 64.058 g/mol, respectively.
3. The number of grams of SO₂ produced can be calculated with the limiting reactant. To find it we need to calculate the number of moles of each reactant.
[tex] n_{{CS_{2}}_{i}} = \frac{m}{M} = \frac{10.5 g}{76.131 g/mol} = 0.138 \:moles [/tex]
[tex] n_{{O_{2}}_{i}} = \frac{41.6 g}{31.998 g/mol} = 1.30 \:moles [/tex]
Knowing that 1 mol of CS₂ reacts with 3 moles of O₂ (eq 3), we have:
[tex] n_{CS_{2}} = \frac{1\: mol\: CS_{2}}{3\: moles\: O_{2}}*1.30 \:moles \:O_{2} = 0.43 \:moles [/tex]
Since we need 0.43 moles of CS₂ to react with O₂ and initially we have 0.138 moles, the limiting reactant is CS₂. We can find now the mass of SO₂ produced.
[tex] n_{SO_{2}} = \frac{2\: moles\: SO_{2}}{1\: mol\: CS_{2}}*0.138 \:moles \:CS_{2} = 0.276 \:moles [/tex]
Now, the mass is:
[tex] m_{SO_{2}} = n_{SO_{2}}*M_{SO_{2}} = 0.276 \:moles*64.058 g/mol = 17.68 g [/tex]
Therefore, the mass of sulfur dioxide produced is 17.68 grams.
You can learn more about limiting reactants here:
I hope it helps you!
