<span>1=H, 2=B, 3=F, 4=A,5=C,6=E, 7=D, 8=G
</span>9: 69Ga=60.12% and 71Ga=39.88%; total=69.797amu
10: 27 27.977 92.23; 28 28.976 4.67; 29 29.974 3.10; abundance =28.07 Silicon
I hope this helps!
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
The limiting reactant is chlorine (Cl2).
<u>Explanation</u>:
Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.
2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.
Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g
= 1.260 g moles of Al
Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g
= 1.10 g moles of Cl2
Dividing each reactant by it's coefficient in the balanced equation obtains:
1.260 moles Al / 2 = 0.63 g moles of Al
1.11 moles Cl2 / 3 = 0.36 g moles of Cl2
The reactant which produces a lesser amount of product is called as limiting reactant.
Here the Limiting reactant is Cl2.