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inysia [295]
1 year ago
15

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO2.

Chemistry
1 answer:
Serhud [2]1 year ago
4 0

Answer:

9.89 x 10²³ molecules H₂S

Explanation:

To find the molecules of H₂S, you need to (1) convert grams S to moles S (via the atomic mass of sulfur), then (2) convert moles S to moles H₂S (via the mole-to-mole ratio from equation coefficients), and then (3) convert moles H₂S to molecules H₂S (via Avogadro's Number). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value.

Atomic Mass (S): 32.065 g/mol

2 H₂S(s) + SO₂(g) -----> 3 S(s) + 2 H₂O(l)

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

79.0 g S           1 mole            2 moles H₂S          6.022 x 10²³ molecules
---------------  x  ---------------  x  ----------------------  x  -------------------------------------  =
                        32.065 g            3 moles S                          1 mole

=  9.89 x 10²³ molecules H₂S

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
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Answer : The enthalpy of neutralization is, 56.012 kJ/mole

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First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

The balanced chemical reaction will be,

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From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

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q = heat absorbed = ?

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T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

Now put all the given values in the above formula, we get:

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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