Answer:
6.19 mol H₂O
Explanation:
First, we<u> convert 70.25 grams of NH₃ into moles</u>, using its<em> molar mass</em>:
Then we <u>convert 4.13 moles of NH₃ into moles of H₂O</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
1. S
+ Sr(OH)
⇒ SrSO4 + 2 H2O is the balanced reaction.
2. 0.034 liters of KI will be required completely react with 2.43 g of Cu(NO3)2.
3. 0.55 liters is the volume in L of a 0.724 M Nal solution contains0.405 mol of NaI.
4. 33.3 ml many mL of 0.300 M NaF would be required to make a 0.0400 M solution of NaF when diluted to 250.0 mL with water
Explanation:
Balance chemical reaction of neutralization:
1. S
+ Sr(OH)
⇒ SrSO4 + 2 H2O
2. Data given:
balance chemical reaction:
2Cu(NO₃)₂(aq) + 4KI(aq) → 2CuI(aq) + I₂(s) + 4KNO₃(aq)
molarity of KI = 0.209 M
mass of Cu(NO₃)₂ = 2.43 grams
number of moles of Cu(NO₃)₂ will be calculated as:
number of moles =
atomic mass of Cu(NO₃)₂ = 187.56 grams/mole
putting the values in the equation,
number of moles=
= 0.0129 moles
2 moles of Cu(NO₃)₂ will react with 4 moles of KI
0.0129 moles will react with x moles of KI
=
x = 0.0258
atomic mass of KI = 166.00 grams/mole
mass = 166.00 x 0.0258
= 4.28 grams or ml is the final volume of KI
so molarity =
so molarity of KI is 0.0258 M, volume is 1 litre.
Using the formula
Minitial x Vinitial = M final Vfinal
V initial =
=
= 34.67 ml 0.034 liters of KI will be required.
3) Data given:
molarity of NaI = 0.724
number of moles of NaI =?
Volume in litres =?
formula used:
molarity =
volume in litres =
= 0.55 liters is the volume
4) Data given:
Initial molarity = 0.3 M
initial volume = ?
final molarity = 0.04 M
final volume diluted by = 250 ml
formula used:
M initial X Vinitial = Mfinal X V final
putting the values in the equation:
Vinitial =
= 33.3 ml of 0.3 M solution will be required.