The percent yield of this reaction is calculated as follows
Mg3N2 + 3H2O =2NH3 + 3Mgo
calculate the theoretical yield,
moles=mass/molar mass
moles Mg3N2= 3.82 g/100g/mol= 0.0382 moles(limiting regent)
moles of H2o= 7.73g/18g/mol = 0.429 moles ( in excess_)
by use of mole ratio between Mg3N2 to MgO which is 1:3 the moles of MgO = 0.0382 x3 = 0.1146 moles
mass =moles x molar mass
the theoretical mass is therefore = 0.1146mole x 40 g/mol = 4.58 grams
The % yield = actual mass/theoretical mass x1000
= 3.60/4.584 x100= 78.5%
Answer: 23 g
the mol of Fe is 11,5/56 = 23/112 (mol)
the mol of CO is 
(mol)
we have:
Fe2O3 + 3CO => 2Fe + 3CO2
23/112 69/112 23/56
=> the mol of Fe is 23/56 mol
=> the mass of Fe is 23/56.56 = 23 (g)
Explanation:
Answer:
A
Explanation:
It is correct please I hope it helps! :)
The part of the atom that takes up the most space is the amount of electrons
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
