mass of PbI₂ = 27.6606 g
<h3>Further explanation</h3>
Given
Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃
28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI
Required
mass of PbI₂
Solution
Balanced equation
Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃
The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product
mol Pb(NO₃)₂ :
= 28 : 331,2 g/mol
= 0.0845
mol NaI :
= 18 : 149,89 g/mol
= 0.12
Limiting reactant : mol : coefficient
Pb(NO₃)₂ : 0.0845 : 1 = 0.0845
NaI : 0.12 : 2 = 0.06
NaI limiting reactant (smaller ratio)
mol PbI₂ based on NaI
= 1/2 x 0.12 = 0.06
Mass PbI₂ :
= 0.06 x 461,01 g/mol
= 27.6606 g
It’s A and D have a good day!!
Answer:
21 mL of KOH
Explanation:
We have to start with the <u>reaction</u> between HCl and KOH:
Now, using the <u>molarity equation</u> we can <u>find the moles</u> of HCl, (we have to keep in mind that we have to use the <u>volume in L</u>, 21 mL=0.021 L)
Now, we can convert from mol of HCl to mol of KOH using the <u>molar ratio</u> of the balanced reaction. (1 mol HCl= 1 mol KOH)
Finally, we have to calculate the <u>volume of KOH</u> using the molarity equation again:
When we <u>convert </u>this value to mL units we will get:
<u>We will need 21 mL of KOH at the equivalence point</u>.
Elements with a plus one charge have the fewest valence electrons among all groups and elements in the periodic table. This includes lithium, soldium, potassium, rubidium, cesium and francium. The number of valence electrons increase from left to right. The highest number of valence electrons include the noble gases.
Explanation:
I'll give you a real answer later need points for test questions sorry