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
There are 1000 mililiters in a liter, so 1000 ml for every liter, you have 5 liters, so:
5L*1000 = 5000 mL
Answer:
Kenma I'll help you! it's B.
Explanation:
I hope I helped you Kenma
Answer: The value of the equilibrium constant Kc for this reaction is 3.72
Explanation:
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For the given chemical reaction:
The expression for is written as:
![K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E3%5Ctimes%20%5BH_2O%5D%5E1%7D%7B%5BHNO_3%5D%5E2%5Ctimes%20%5BNO%5D%5E1%7D)
Thus the value of the equilibrium constant Kc for this reaction is 3.72
