When a star dies which of these celestial objects is it most likely to help create ?
2 answers:
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Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
<u>Answer:</u> The theoretical yield of solid lead comes out to be 5.408 grams.
<u>Explanation:</u>
To calculate the moles, we use the following equation:
- <u>Moles of Lead nitrate:</u>
Given mass of lead nitrate = 8.65 grams
Molar mass of lead nitrate = 331.2 g/mol
Putting values in above equation, we get:
- <u>Moles of Aluminium:</u>
Given mass of aluminium = 2.5 grams
Molar mass of aluminium = 27 g/mol
Putting values in above equation, we get:
For the given chemical reaction, the equation follows:
By Stoichiometry:
3 moles of lead nitrate reacts with 2 moles of aluminium
So, 0.0261 moles of lead nitrate are produced by = of aluminium.
As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.
Lead nitrate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
3 moles of lead nitrate are produces 3 moles of lead metal.
So, 0.0261 moles of lead nitrate will produce = of lead metal.
- Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:
Molar mass of lead = 207.2 g/mol
Putting values in above equation, we get:
Mass of lead = 5.408 grams
Hence, the theoretical yield of solid lead comes out to be 5.408 grams.