Calculate the molar concentration of the Cl⁻ ions in 0.92 M MnCl2(aq).
1 answer:
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Answer:
A. fluorine, 1.79 moles
Explanation:
Given parameters:
Mass of carbon = 87.7g
Mass of fluorine gas = 136g
Unknown:
The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?
Solution:
Equation of the reaction:
C + 2F₂ → CF₄
let us find the number of the moles the given species;
Number of moles =
C; molar mass = 12;
Number of moles = = 7.31moles
F; molar mass = 2(19) = 38g/mol
Number of moles = = 3.58moles
So;
From the give reaction:
1 mole of C requires 2 moles of F₂
7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles
But we have 3.58 moles of the F₂;
Therefore, the reactant in short supply is F₂ and it is the limiting reactant;
So;
2 moles of F₂ will produce mole of CF₄
3.58 moles of F₂ will then produce = 1.79moles of CF₄
Answer:
Explanation:
From the given information:
mass of silver chloride AgCl = 11.89 g
molar mass of AgCl = 143.37 g/mol
We know that:
number of moles = mass/molar mass
∴
number of moles of AgCl = 11.89 g/ 143.37 g/mol
number of moles of AgCl = 0.0829 mol
The chemical equation for the mineral called trona is:
when being reacted with hydrochloric acid, we have:
One mole of NaCl formed from one mole of trona sample = 0.0829 moles of AgCl
i.e. 0.0829 moles of NaCl can be formed from AgCl
mass of trona sample = number of moles × molar mass
mass of trona sample = 0.0829 × 226
mass of trona sample = 18.735 g
The mass in the percentage of NaHCO₃ = mass of NaHCO₃/ mass of trona
The mass in the percentage of NaHCO₃ = 6.93/18.735
The mass in the percentage of NaHCO₃ = 0.36989
The mass in the percentage of NaHCO₃ = 36.99%
Nonetheless, a 6.78 g samples manufactured from sodium carbonate in pure 100%
∴
6.78 g sample manufactured from Na₂CO₃ is purer.