Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
Sulfuric acid reacts violently with alcohol and water to release heat. It reacts with most metals, particularly when diluted with water, to form flammable hydrogen gas, which may create an explosion hazard. ... Hazardous decomposition products are as follows: sulfur dioxide, sulfur trioxide, and sulfuric acid fumes.
Answer is: 0.102 moles of HCl would react.
Balanced chemical reaction:
2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).
V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.
V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.
c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.
n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).
n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.
n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.
From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.
n(HCl) = 2 · n(Sr(OH)₂).
n(HCl) = 2 · 0.0051 mol.
n(HCl) = 0.0102 mol; amount of the hydrochloric acid.
Answer:
b. have the same kind number of complete shells
