Na2CO3 Na= 23*2=46 C=12 O=16*3=48. Na2CO3=106. 46/106*0.175=0=403
<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
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