Answer is: A. 1.81 mol.
Balanced chemical reaction: FeCl₂ + 2KOH → Fe(OH)₂ + 2KCl.
n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.
n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.
From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.
n(Fe(OH)₂) = n(KOH) ÷ 2.
n(Fe(OH)₂) = 3.62 mol ÷ 2.
n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.
Answer;
1.6 kg.
Solution;
The density is 1.36 g/ml;
The volume is 1.25 qt
However; 1 qt = 946.35 ml
Mass is given by; density × volume;
= 1.25 qts × 946.25 ml/qt × 1.36 g/ml =1609 g
but; 1 kg = 1000 g
Hence the mass = 1609/1000 = 1.609 Kg or 1.61 (sig figs)
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
Answer:
Explanation:
<h2>Before you use log you have to multiply the answer of 8.66 mol by 6.02⋅1023 by 3 because the answer of 8.66 mol by 6.02⋅1023 gives you the number of ammonia molecules and in each molecules there are 3 hydrogen atoms present.
</h2><h2>
</h2><h2>So it is:
</h2><h2>
</h2><h2>8.66⋅6.02⋅1023=5.21332⋅1024
</h2><h2>5.21332⋅1024⋅3=1.563996⋅1025
</h2><h2>log(1.563996⋅1025)=25.19
</h2>
