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satela [25.4K]
3 years ago
5

Solid silver chloride, AgCI(s), decomposes to its elements when exposed to sunlight Write a balanced equation for this reaction

with correct formulas for the products. What is the coefficient in front of AgCI(s) when the equation is properly balanced? 01 03 02 04
Chemistry
1 answer:
Mrac [35]3 years ago
8 0

Answer: The coefficient in front of AgCl when the equation is properly balanced is 2.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Decomposition of silver chloride is represented as:

2AgCl(s)\rightarrow 2Ag(s)+Cl_2(g)

Thus the coefficient in front of AgCl when the equation is properly balanced is 2.

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When heated, lithium reacts with nitrogen to form lithium nitride: 6Li(s) + N2(g) → 2Li3N(s) What is the theoretical yield of Li
anyanavicka [17]

Answer:

The % yield of the reaction = 27.5 %

Explanation:

Step 1: Data given

Mass of Li = 12.7 grams

Mass of N2 = 34.7 grams

Actual yield of Li3N = 5.85 grams

Molar mass of  Lithium = 6.94 g/mol

Molar mass of N2 = 28 g/mol

Molar mass of LI3N = 34.83 g/mol

Step 2: The balanced equation:

6Li(s) + N2(g) → 2Li3N(s)

Step 3: Calculate moles of Lithium

Moles Li = mass Li / Molar mass Li

Moles Li = 12.7 grams / 6.94 g/mol

Moles Li = 1.83 moles

Step 4: Calculate moles of N2

Moles N2 = 34.7 g/ 28 g/mol

Moles N2 = 1.24 moles

Step 5: Limiting reactant

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

Lithium is the limiting reactant. It will completely be consumed (1.83 moles).

N2 is in excess. There will be consumed 1.83 / 6 = 0.305 moles

There will remain 1.24 - 0.305 = 0.935 moles

Step 6: Calculate moles of Li3N

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

For 1.83 moles Li, we'll have 1.83/3 = 0.61 moles of Li3N

Step 7: Calculate mass of Li3N

Mass Li3N =moles LI3N * Molar Mass LI3N

Mass Li3N = 0.610 moles * 34.83 g/mol

Mass Li3N = 21.2463 grams = Theoretical yield

Step 8: Calculate % yield

% yield = actual yield / theoretical yield

% yield = (5.85 / 21.2463)*100% = 27.5%

The % yield of the reaction = 27.5 %

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3 years ago
How many grams of potassium (K) contain 5.11 x 10^22 atoms of potassium?
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The atomic mass of K is 39

from Avogadro's law

39g of K contains 6.02x10^23 atoms

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making X the subject of the formula

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