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11111nata11111 [884]
3 years ago
6

Silver nitrate solution reacts with calcium chloride solution according to the equation: 2 AgNO3 + CaCl2 → Ca(NO3)2 + 2 AgCl All

of the substances involved in this reaction are soluble in water except silver chloride, AgCl, which forms a solid (precipitate) at the bottom of the flask. Suppose we mix together a solution containing 6.97 g of AgNO3 and one containing 6.39 g of CaCl2. What mass, in grams, of AgCl is formed?
Chemistry
1 answer:
HACTEHA [7]3 years ago
5 0

Answer:

14.33 g

Explanation:

Solve this problem based on the stoichiometry of the reaction.

To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and  finally calculate the mass of AgCl.

                      2 AgNO₃   + CaCl₂    ⇒   Ca(NO₃)₂   + 2 AgCl

mass, g                  6.97        6.39                                     ?

MW ,g/mol         169.87      110.98                                  143.32

mol =m/MW           0.10         0.06                                    0.10

From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :

0.10 mol x 143.32 g/mol = 14.33 g

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p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}

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===============

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We can use the Ideal Gas Law.

<em>pV = nRT</em>                Divide both sides by <em>RT</em> and switch

<em>n</em> = (<em>pV</em>)/(<em>RT</em>)

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R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 24.1 °C                                                              Convert to kelvins

T = (24.1 + 273.15 ) K = 297.25 K                          Insert the values

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===============

3)<em> Moles of metal </em>

The partial chemical equation is

M + … ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.001 014 × 1/1                          Do the operations

Moles of M = 0.001 014 mol M

===============

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===============

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The metal with the closest atomic mass is Mg (24.305 g/mol).

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