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LekaFEV [45]
2 years ago
15

What mass of ag2co3 could you produce from 12.7 g agno3 assuming that it is the limiting reagent?

Chemistry
1 answer:
zhenek [66]2 years ago
7 0

10.3 g of  Ag_{2} CO_{3} will produce from 12.7 g AgNO_{3}.

In simple terms, a limiting reagent is a reactant that is completely used up in the reaction. 

It is also referred to as a limiting reactant or limiting agent.

Now, according to the question,

Since it is a limiting reagent, it will react fully. 

The mass of AgCO_{3} that will be produced will be:-

Equivalent mole of AgNO_{3} = Equivalent weight of Ag_{2} CO_{3} = 12.7/169.97 = x*2/274x = 10.3 g. 

Hence, 10.3 g of Ag_{2} CO_{3} will be produced.

It is used to restrict the reaction.

It tells you the estimated amount of compound to be used.

It brings quantitative understanding to chemical reactions.

More information on limiting reagents can be found here :

brainly.com/question/11848702

#SPJ4

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Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo
Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)  

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess.  There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

3 0
3 years ago
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Answer:

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Explanation:

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What is the volume of a 5.30g piece of aluminum? The density for aluminum is 2.70gmL.
Blizzard [7]

The volume of the piece of aluminum is 1.96 mL

Explanation:

Density is the relationship of the mass of a substance and its volume.

In this case, the mass of aluminum is 5.30 g and the density is 2.70 g/mL

The formula to apply is;

D=M/V where D is density in g/mL, M is mass in g and V is volume in mL

2.70=5.30/V

V=5.30/2.70 =1.96 mL

Learn More

Density of a substance:brainly.com/question/12605423

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