The reaction of C4H8 and Br2 to yield C4H2Br2 represents
1 answer:
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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
Answer:
The mass of copper(II) sulfide formed is:
= 81.24 g
Explanation:
The Balanced chemical equation for this reaction is :
given mass= 54 g
Molar mass of Cu = 63.55 g/mol
Moles of Cu = 0.8497 mol
Given mass = 42 g
Molar mass of S = 32.06 g/mol
Moles of S = 1.31 mol
Limiting Reagent :<em> The reagent which is present in less amount and consumed in a reactio</em>n
<u><em>First find the limiting reagent :</em></u>
1 mol of Cu require = 1 mol of S
0.8497 mol of Cu should require = 1 x 0.8497 mol
= 0.8497 mol of S
S present in the reaction Medium = 1.31 mol
S Required = 0.8497 mol
S is present in excess and <u>Cu is limiting reagent</u>
<u>All Cu is consumed in the reaction</u>
Amount Cu will decide the amount of CuS formed
1 mole of Cu gives = 1 mole of Copper sulfide
0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide
= 0.8497
Molar mass of CuS = 95.611 g/mol
Mass of CuS = 0.8497 x 95.611
= 81.24 g