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50 grams of CaCo3 is allowed to react with 73.5 grams of h3po4 calculate amount of Ca3(Po4)2 formed?

50 grams of CaCo3 is allowed to react with 73.5 grams of h3po4 calculate amount of Ca3(Po4)2 formed?

Grade:11

4 Answers

DEEPTHI JANGA
51 Points
7 years ago
good evening yasik guptha,
in my point of veiw,
balanced equation- 3CaCo3+2H2Po4 gives Ca3(Po4)2+3Co2+3H2O  so
210g of CaCo3 and 196 H3Po4 gives 310g of Ca3(Po4)
According to problem,
50gof CaCo3 and 73.5 gH3Po4 gives x g of  Ca3(pO4) then ‘
x=(50+73.5)*310/210+196
i.e. equal to 75.66205g
Kshitij nandwana
24 Points
6 years ago
I th
anonymous
11 Points
5 years ago
applying poac...
let ca be conserved..
so we have
1×moles of caco3=3×moles of ca3(po4)3
1/2=3×moles of ca3(po4)3
moles=1/6
now wt=1/6×310
Aashay
15 Points
5 years ago
3 mole of CaCO3 -: 1 mole Ca3(PO4)2
1 mole of CaCO3 -: 1/3 mole Ca3(PO4)2
1/2 mole CaCO3 -: 1/6 mole Ca3(PO4)2
So 1/6 moles of Ca3(PO4)2 will have mass 51.666 g

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