1)
Molar mass of Ca3(PO4)2,
MM = 3*MM(Ca) + 2*MM(P) + 8*MM(O)
= 3*40.08 + 2*30.97 + 8*16.0
= 310.18 g/mol
mass(Ca3(PO4)2)= 127.0 g
use:
number of mol of Ca3(PO4)2,
n = mass of Ca3(PO4)2/molar mass of Ca3(PO4)2
=(1.27*10^2 g)/(3.102*10^2 g/mol)
= 0.4094 mol
Molar mass of H2SO4,
MM = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol
mass(H2SO4)= 71.9 g
use:
number of mol of H2SO4,
n = mass of H2SO4/molar mass of H2SO4
=(71.9 g)/(98.09 g/mol)
= 0.733 mol
Balanced chemical equation is:
Ca3(PO4)2 + 3 H2SO4 ---> 2 H3PO4 + 3 CaSO4
1 mol of Ca3(PO4)2 reacts with 3 mol of H2SO4
for 0.4094 mol of Ca3(PO4)2, 1.228 mol of H2SO4 is required
But we have 0.733 mol of H2SO4
so, H2SO4 is limiting reagent
we will use H2SO4 in further calculation
Molar mass of H3PO4,
MM = 3*MM(H) + 1*MM(P) + 4*MM(O)
= 3*1.008 + 1*30.97 + 4*16.0
= 97.994 g/mol
According to balanced equation
mol of H3PO4 formed = (2/3)* moles of H2SO4
= (2/3)*0.733
= 0.4887 mol
use:
mass of H3PO4 = number of mol * molar mass
= 0.4887*97.99
= 47.89 g
Answer: 47.9 g
Only 1 question at a time please
Suppose the reaction Ca3(PO4)2 + 3H2SO4 - 3CaSO4 + 2H3PO4 is carried out starting with 127...
Suppose the reaction Ca3(PO4)2 + H2SO4 - CaSO4 + H3PO4 is carried out starting with 149 g of Ca3(PO4)2 and 86.9 g of H2SO4. How much phosphoric acid will be produced? ООО 86.8 g 57.98 235.9 g 130.28 94.18 C
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