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:
Explanation:
Hello,
In this case, as the atomic mass of coppe is 63.546 g/mol, with the given mass with can compute the moles as shown below:
Best regards.
Answer:
4.1 moles
Explanation:
Applying
PV = nRT................ equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT.............. Equation 2
From the question,
Given: V = 35 L , P = 2.8 atm, T = 15 °C = (15+273) = 288 K, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (35×2.8)/(0.083×288)
n = 4.1 moles
