Answer:
Volume required = 0.327 L
Explanation:
Given data:
Volume in L = ?
Molarity of solution = 1.772 M
Mass of BaCl₂ = 123 g
Solution:
First of all we will calculate the number of moles of BaCl₂,
Number of moles = mass/molar mass
Number of moles = 123 g/ 208.23 g/mol
Number of moles = 0.58 mol
Now, given problem will solve by using molarity formula.
Molarity = number of moles / volume in L
1.772 M = 0.58 mol / Volume in L
Volume in L = 0.58 mol / 1.772 M
Volume in L = 0.327 L
The answer above is correct (I took a test on this)
Answer:
0.55 mol Au₂S₃
Explanation:
Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.
1. Gather all the information in one place:
M_r: 34.08
Au₂S₃ + … ⟶ 3H₂S + …
m/g: 56
2. Calculate the moles of H₂S
Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)
= 1.64 mol H₂S
3. Calculate the moles of Au₂S₃
The molar ratio is 1 mol Au₂S₃/3 mol H₂S.
Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)
= 0.55 mol Au₂S₃
