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LenaWriter [7]
3 years ago
7

For the chemical reaction CaI2+2AgNO3-> 2AgI+Ca(NO3)2 how many moles of silver iodide will be produced from 205g of calcium i

odide
Chemistry
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃     →      2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

                     CaI₂         :           AgI

                         1           :             2

                       0.7         :           2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

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Answer:

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Explanation:

Given:

Volume of NH_{4} HS    V = 2 lit

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The reaction in which NH_{3} is produced

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Here equal moles of NH_{3} and H_{2}S is formed.

From the formula of equilibrium constant,

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So in 2 L no. moles of NH_{3} = 0.061 \times 2 = 0.122 moles.

So mass of 0.122 mole of NH_{3} is = 0.122 \times 17 = 2.074 g

Therefore, the mass of NH_{3} in the container is 2.074 gram

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