0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess
P4)
<h3>How to calculate moles?</h3>
The balanced chemical equation is
The mass of clorine is m(
) = 58.0 g
The amount of clorine is n() = m()/M() = 58/70.906 = 0.817 mol
The stoichiometric reaction,shows that
10 moles of yield 4 moles of 
;
0.817 of yield x moles of
n() = 4*0.817/10 = 0.3268 mol
To know more about stoichiometric reaction, refer:
brainly.com/question/14935523
#SPJ9
I thought the answer is d
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
To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons.
