Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L</span>
Start by adding the numbers then divide
Answer: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^4
Explanation:
I suggest looking at the electron configuration chart, it has really helped me a lot :)
Answer:
Sr 2+(aq) + SO42-(aq) → SrSO4(s)
Explanation:
<u>Step 1</u>: Write a properly balanced equation with states:
K2SO4(aq) + Srl2(aq) → 2KI(aq) + SrSO4(s)
<u>Step 2</u>: write the full ionic equation with states. Remember to keep molecules intact. Only states (aq) will dissociate, (s) will not dissociate
. This means SrSO4 won't dissociate.
2K+(aq) + SO42-(aq) + Sr 2+(aq) + 2I-(aq) → 2K+(aq) + 2I-(aq) + SrSO4(s)
<u>Step 3</u>: Balanced net ionic equation
Sr 2+(aq) + SO42-(aq) → SrSO4(s)
