About 4 C to for water density until it is cooled
Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass
moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles
moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4
Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles
mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams
Answer:
12 moles of water will be produced
Explanation:
Given data:
Number of moles of NH₃ = 8.00 mol
Number of moles of O₂ = 14.0 mol
Number of moles of H₂O produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ +6H₂O
Now we will compare the moles of reactant with product.
NH₃ : H₂O
4 : 6
8 : 6/4×8 = 12
O₂ : H₂O
7 : 6
14 : 6/7×14 = 12
12 moles of water will be produced.
?? Is that the whole question?
