I believe B is the correct answer
<h3><u>Answer;</u></h3>
321.8 g CaF2
321.5 g Al2(CO3)3
<h3><u>Explanation;</u></h3>
The equation for the reaction is;
3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2(CO3)3
Number of moles of CaCO3 will be;
=(412.5 g CaCO3) / (100.0875 g CaCO3/mol)
= 4.12139 mol CaCO3
Number of moles of AlF3 will be;
= (521.9 g AlF3) / ( 83.9767 g AlF3/mol)
= 6.21482 mol AlF3
But;
4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.
Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.
Therefore;
Mass of CaF2 will be;
(4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2
Mass of Al2(CO3)3 on the other hand will be;
(4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2(CO3)3/mol) = 321.5 g Al2(CO3)3
Answer:
8
Explain
H: Ignoring the coefficient, we know there's 5+3=8 atoms
Answer:
M = 1.18 mol/L
Explanation:
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = moles ,
w = given mass ,
m = molecular mass .
From the information of the question ,
w = 78.3 g
As we known ,
molecular mass of 
is 189.36 g/mol
m = 189.36 g/mol
moles can be calculated as -
n = 0.413 mol
MOLARITY -
Molarity of a substance , is the number of moles present in a liter of solution .
M = molarity ( unit = mol / L or M )
V = volume of solution in liter ( unit = L ),
n = moles of solute ( unit = mol ),
From the question ,.
V = 350 mL
Since , 1 mL = 10⁻³ L
V = 0.350 L
n = 0.413 mol
Molarity can be calculated as -
M = 1.18 mol/L
