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dexar [7]
3 years ago
8

What is the amount of diamine silver that can be formed when 10.00 g AgCl is mixed with 1.00 L of 0.100 M NH3?

Chemistry
1 answer:
Murljashka [212]3 years ago
6 0

The amount of diamine silver chloride = 8.87 g

<h3>Further explanation</h3>

Given

10 g AgCl

1.00 L of 0.100 M NH3

Required

the amount of diamine silver

Reaction

AgCl + 2 NH₃ → [Ag(NH₃)₂]Cl

mol AgCl :

= mass : MW

= 10 g : 143,32 g/mol

= 0.0698

mol NH₃ :

= M x V

= 0.1 x 1

= 0.1

NH₃ as a limiting reactant

mol [Ag(NH₃)₂]Cl based on NH₃ :

= 1/2 x mol NH₃

= 1/2 x 0.1

= 0.05

Mass diamine silver :

= 0.05 x  177.3822 g/mol

= 8.87 g

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2.50g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 12
Art [367]

Answer:

                      Molecular Formula  =  C₁₀H₈

Explanation:

Step 1: <u>Calculating Moles for each Element:</u>

C  =  Mass of CO₂ × (1 mol of CO₂÷ M.Mass of CO₂) × (1 mol of C ÷ 1 mol of CO₂)

C  =  8.60 g × (1 mol of CO₂÷ 44.011 g.mol⁻¹) × (1 mol of C ÷ 1 mol of CO₂)

C =  0.1954 mol

H  =  Mass of H₂O × (1 mol of H₂O ÷ M.Mass of H₂O) × (2 mol of H ÷ 1 mol of H₂O)

H  = 1.41 g × (1 mol of H₂O ÷ 18.106 g.mol⁻¹) × (2 mol of H ÷ 1 mol of H₂O)

H =  0.1557 mol

Step 2: <u>Calculate the Smallest whole number ratio as,</u>

                                 C                                                        H

                              0.1954                                               0.1557

                      0.1954/0.1557                                  0.1557/0.1557

                               1.25                                                       1

Multiply ratio by 4,

                                 5                                                         4

Result:

          Empirical Formula of Propane is C₅H₄

Step 3: <u>Calculate Molecular Formula:</u>

Molecular formula is calculated by using following formula,

                    Molecular Formula  =  n × Empirical Formula  ---- (1)

Also, n is given as,

                     n  =  Molecular Weight / Empirical Formula Weight

Molecular Weight  =  128.0 g.mol⁻¹

Empirical Formula Weight  =  5 (C) + 4 (H) =  64 g.mol⁻¹

So,

                     n  =  128.0 g.mol⁻¹ ÷ 64 g.mol⁻¹

                     n  =  2

Putting Empirical Formula and value of "n" in equation 1,

                    Molecular Formula  = 2 × C₅H₄

                    Molecular Formula  =  C₁₀H₈

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