Question

If I have 2 samples of limestone, sample 1 weighing 0.732g and sample 2 weighing 0.825g. Both were dissolve do in15ml of 3.00M of HCL. 1.5g of (NH4)2SO4 was added to each sample. Then 20ml of 2.00M NH3OH was added to each sample. How do I calculate the moles of CaCO3 in limestone AND the mass of CaCO3 in limestone ?

Answer 1

WE WILL FIRST CALCULATE THE Moles of HCL ,

MOL WT OF HCL=36.5, Then we will calculate , 36.5 grams in 1000 ml=1 molar,then for 15 ml how many grams of hcl?, by cross multiplying, 5*36.5=547.5/1000=0.5475 grams (for 1 molar),

then for 3 molar =3*0.5457=1.6407 grams of HCL./36.5=0.0455 MOLES OF HCL. IN 15 ML SOLUTION.

For reaction of caco3 from limestone 2 moles of HCL IS REQUIRED TO Dissolve 1 mol of caco3.

CaCo3+2 HCL---------->CaCl2+ H2O,

0.045 MOLES HCL/2= 0.02278 MOLES OF CaCO3.

which then dissolved to 1.5 g ammonium sulphate moles=1.5/132.14=0.011 moles means only half mole is carry forward to caso4 and then 20 ml ammonium hydroxide 2 molar means,

35 grams in 1000 ml=1 molar,then for 20 ml how many grams of NH4OH?, by cross multiplying, 20*35=700/1000=0.7 grams (for 1 molar), THEN for 2 molar 2*0.7=1.40 grams/35 mol wt= 0.04 mol,

final product is Ca(OH)2, MOL WT = 74.098*0.11 MOLES WILL BE FORMED., NOW WECAN CALCULATE THE THEROTICAL YIELD,

SAMPLE 1=GMS OF REACTANT*MOLWT OF PRODUCT/MOLWT OF REACTANT

=0.732*74.09=54.23/100.08=0.5419 GRAMS, (MOL WT OF CaCO3=100.08),

0.5419/74.09=0.0073 moles Ca(OH)2, THIS MEANS THAT, 0.0073 moles caco3 is present in limestone. mass of caco3=0.0073*100.08=0.732 grams)

sample 2=GMS OF REACTANT*MOLWT OF PRODUCT/MOLWT OF REACTANT

=0.825*74.09=61.12/100.08=0.610 grams,

0.610 /74.09=0.0082 moles of Ca(OH)2, MEANS( 0.0082*100.08=0.825 GRAMS OF CACO3)