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Orlov [11]
3 years ago
12

According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed?

Chemistry
1 answer:
julia-pushkina [17]3 years ago
8 0

Answer:

Mass = 100.8 g

Explanation:

Given data:

Mass of sulfur formed = ?

Mass of water formed = 37.4 g

Solution:

Chemical equation:

2H₂S + SO₂      →   3S + 2H₂O

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 37.4 g/ 18 g/mol

Number of moles = 2.1 mol

Now we will compare the moles of water and sulfur.

          H₂O         :            S

            2             :          3

           2.1            :         3/2×2.1 = 3.15

Mass of sulfur:

Mass = number of moles × molar mass

Mass = 3.15 mol × 32 g/mol

Mass = 100.8 g

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4 0
2 years ago
using the soubility curve what is the solubilityof nh4cl in 10 mL of water at a temperature of 60 degrees Celsius​
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Answer:

Please, see attached two figures:

  • The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.

  • The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.

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Explanation:

The red  arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>

From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.

Assuming density 1.0 g/mol for water, 10 mL of water is:

            10mL\times 1.0g/mL=10g

Thus, the solutibily is:

      10gWater\times 55gNH_4Cl/100gWater=5.5gNH_4Cl

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Explanation:

Hope this helps!

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