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4 years ago

How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride?

Chemistry

1 answer:

Mumz [18]4 years ago

6 0

Answer:

4.21 grams of silver chloride are produced

Explanation:

Let's follow the reaction:

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

Ratio is 2:2 so 2 moles of nitrate produce, 2 mol of chloride.

Mol = Mass / Molar mass → 5g / 169.87 g/m = 0.0294 moles

So, 0.0294 moles of Silver chloride are produced

Molar mass AgCl = 143.32 g/m

Molar mass . mol = Mass → 143.32 g/m . 0.0294 mol = 4.21 grams

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Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

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$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

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$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

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