Question

100gm of Sample of caco3 reacts with
98grmsе оf H3PO4 -
calculate
a) no of grams of Ca 3 (PO4)2 that could
be produced
b) the no of grams of excess reagent that
will remain unreacted​

Answer 1

Answer:

Explanation:

first find limiting reagent

for finding limiting reagent find mole/stchiometric coefficient

in that which is less that will be limiting reagent

and we know that all depends on limiting reagent

moles of caco3 = 100/100=1

moles of h3po4 = 98/98=1

both are limiting here

if case gets failed find equivalent

less equivalent specie is limiting reagent

equivalent of caco3  = moles*f=1*2=2

equivalent of h3po4 = moles*f = 1*2=2

and by this case also both are limiting reagent

that means this is equivalence point

both gets neutralised and form product of same equivalent

equivalent of ca3(po4)2=2

w/310*6 = 2

w=620/6

w=103.34 gm

2)no. of grams of excess reagent that remain unreacted = 0