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nexus9112 [7]
3 years ago
5

Hydrogen sulfide burns form sulfur dioxide:

Chemistry
1 answer:
Helga [31]3 years ago
7 0

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of H_2S

\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles

2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)    \Delta H=-1036kJ

According to stoichiometry :

2 moles of H_2S on burning produces = 1036 kJ

Thus 0.78 moles of H_2S on burning produces =\frac{1036kJ}{2}\times 0.78=404.04

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

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Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: V_{1} = 21.35 mL,        M_{1} = 0.150 M

V_{2} = 25.0 mL,           M_{2} = ?

Formula used to calculate molarity of H_{2}SO_{4} is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M

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= 46.36 mL  (1 mL = 0.001 L)

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Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol

Thus, we can conclude that there are 0.006 moles of acid in the flask.

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