Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles =
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
HBr reacts with LiOH and forms LiBr and H₂O as the products. The balanced reaction is
LiOH(aq) + HBr(aq) → LiBr(aq) + H₂O(l)
Molarity (M) = moles of solute (mol) / volume of the solution (L)
Molarity of LiOH = 0.205 M
Volume of LiOH = 29.15 mL = 29.15 x 10⁻³ L
Hence,
moles of LiOH = molarity x volume of the solution
= 0.205 M x 29.15 x 10⁻³ L
= 5.97575 x 10⁻³ mol
The stoichiometric ratio between LiOH and HBr is 1 : 1.
Hence,
moles of HBr in 25.0 mL = moles of LiOH added
= 5.97575 x 10⁻³ mol
Hence, molarity of HBr = 5.97575 x 10⁻³ mol / 25.00 x 10⁻³ L
= 0.23903 M
≈ 0.239 M
Hence, the molarity of the HBr is 0.239 M.