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%
Answer:
The molecules of solid CO2 are much closer together than the molecules of CO2 gas.
Answer:
Kc = 0.5951 (4 sig. figs.)
Explanation:
For A + B ⇄ C + D at standard thermodynamic conditions (298K, 1atm)
ΔG = ΔG° + R·T·lnQ => 0 = ΔG° + R·T·lnKc => ΔG° = - R·T·lnKc
=> lnKc = - ΔG°/R·T
ΔG° = +12.86 Kj/mol
R = 8.314 Kj/mol·K
T = 298K
lnKc = - (+12.86Kj) / (8.314Kj/mol·K)(298K) = - 0.519 mol⁻¹
Kc = e⁻⁰°⁵¹⁹ mol⁻¹ = 0.5957 mol⁻¹ (4 sig. figs.)
The mass of sodium chloride at the two parts are mathematically given as
- m=10,688.18g
- mass of Nacl(m)=39.15g
<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>
Generally, the equation for ideal gas is mathematically given as
PV=nRT
Where the chemical equation is
F2 + 2NaCl → Cl2 + 2NaF
Therefore
1.50x15=m/M *(1.50*0.0821)
1-50 x 15=m/58.5 *(1.50*0.0821)
m=10,688.18g
Part 2
PV=m'/MRT
1*15=m'/58.5*0.0821*273
m'=39.15g
mass of Nacl(m)=m'=39.15g
Read more about Chemical Reaction
brainly.com/question/11231920
#SPJ1
Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
Best regards.
