Answer:
4.08 × 10⁻³
Explanation:
Step 1: Write the balanced reaction at equilibrium
NH₄I(s) ⇄ NH₃(g) + HI(g)
Step 2: Calculate the equilibrium constant
The equilibrium constant (K) is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. Only gases and aqueous species are included.
![K = [NH_3] \times [HI] = 4.34 \times 10^{-2} \times 9.39 \times 10^{-2} = 4.08 \times 10^{-3}](https://tex.z-dn.net/?f=K%20%3D%20%5BNH_3%5D%20%5Ctimes%20%5BHI%5D%20%3D%204.34%20%5Ctimes%2010%5E%7B-2%7D%20%20%5Ctimes%209.39%20%5Ctimes%2010%5E%7B-2%7D%20%3D%204.08%20%5Ctimes%2010%5E%7B-3%7D)
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
The molar mass of copper is 63.55 g/mol
The hallogens chloride with br
