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QUESTION 6

1. What is the maximum amount (in grams) of sulfur dioxide that can be produced by the reaction of 1380 g

of dihydrogen sulfide with 6890 g of oxygen according to the balanced equation below?

2 H2S(g) + 3 O2(g) + 2 H2O(g) + 2 SO2(g)

Enter the numeric value in the box without units.

Respuesta :

Answer:

Explanation:

2 H₂S(g)   +   3O₂(g)  =  2 H₂O(g)   +   2SO₂(g)

2 moles           3 moles                         2 moles

1380 g of H₂S = 1380 / 34 = 40.588 moles of H₂S

6890 g of oxygen  = 6890 / 32 = 215.31 moles of oxygen

Here H₂S is limiting reagent .

2 mole of H₂S  reacts to give 2 moles of sulfur dioxide .

40.588 moles of H₂S will react to give 40.588 moles of sulfur dioxide.

sulfur dioxide formed = 40.588 moles

= 40.588 x 64 = 2597.63 g of sulfur dioxide.

The maximum amount of sulfur dioxide is 2597.63 g of sulfur dioxide.

Calculation of maximum amount:

Since

2 H₂S(g)   +   3O₂(g)  =  2 H₂O(g)   +   2SO₂(g)

So,

2 moles           3 moles                         2 moles

Now

= 1380 g of H₂S

= 1380 / 34

= 40.588 moles of H₂S

Now

= 6890 g of oxygen

= 6890 / 32

= 215.31 moles of oxygen

Here H₂S represents limiting reagent .

2 moles of H₂S  reacts to provide 2 moles of sulfur dioxide .

Now

40.588 moles of H₂S will react to provide 40.588 moles of sulfur dioxide.

Also,

sulfur dioxide formed = 40.588 moles

So,

= 40.588 x 64

= 2597.63 g of sulfur dioxide.

Learn more about reaction here: https://brainly.com/question/21091465

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