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Radda [10]
3 years ago
13

15.50 g of NH4Cl reacts with an excess of AgNO3. In the reaction 35.50 g AgCl is produced. what is the theoretical yield of AgCl

calculated in grams?
Chemistry
1 answer:
Readme [11.4K]3 years ago
8 0

Answer:

The answer to your question is 41.6 g of AgCl

Explanation:

Data

mass of NH₄Cl = 15.5 g

mass of AgNO₃ = excess

mass of AgCl = 35.5 g

theoretical yield = ?

Process

1.- Write the balanced chemical reaction.

              NH₄Cl  +  AgNO₃   ⇒   AgCl  +  NH₄NO₃

2.- Calculate the molar mass of NH₄Cl and AgCl

NH₄Cl = 14 + 4 + 35.5 = 53.5 g

AgCl = 108 + 35.5 = 143.5 g

3.- Calculate the theoretical yield

                 53.5 g of NH₄Cl -------------------- 143.5 g of AgCl

                  15.5 g of NH₄Cl  -------------------    x

                         x = (15.5 x 143.5) / 53.5

                         x = 2224.25 / 53.5

                         x = 41.6 g of AgCl

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4 0
3 years ago
If the equation: FeCl3+O2-->Fe2O3+Cl2, how many moles of chlorine gas can be produced if 4 moles of FeCl3 react with 4 moles
BlackZzzverrR [31]

Answer:

6 moles of Cl2

Explanation:

First, the equation has to be balanced, which makes it 4 FeCl3 + 3 O2 --> 2 Fe2O3 + 6 Cl2

Using this information, we can see that one mole of O2 will not be present in the reaction. Since four moles of FeCl3 are needed to react in the equation, which would produce six moles of Cl2, and only four moles of FeCl3 are present, six moles of Cl2 would be produced.

4 0
2 years ago
If 16.5g of c6h14o2 are reacted vwith .499 mol of o2. How many moles of co2 should be produced
Fed [463]

Answer:

moles of carbon dioxide produced are 410.9 mol.

Explanation:

Given data:

Mass of C₆H₁₄O₂ = 16.5 g

Moles of O₂ = 499 mol

Moles of CO₂ = ?

First of all we will write the balance chemical equation.

2C₆H₁₄O₂  +  17O₂  →   14CO₂  +  12H₂O

moles of C₆H₁₄O₂  = mass × molar mass

moles of C₆H₁₄O₂ =  16.5 g × 118 g/mol

moles of C₆H₁₄O₂ = 1947 mol

Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.

                 O₂      :     CO₂

                 17       :      14

                499     :      14/17× 499 = 410.9 moles

          C₆H₁₄O₂   :   CO₂

                    2    :      14

                 1947 :     14/2× 1947 =  13629 moles

Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.

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