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Please do not make any marks on this exam. 5. The following compounds react Ca3(

ID: 697769 • Letter: P

Question

Please do not make any marks on this exam. 5. The following compounds react Ca3(PO4)2 + H2SO4 How much acid product is formed if you start with 103g of calcium phosphate and 75.0 g of H:SO4? D) 32.5 g E) 97.6g A) 74.9 g B) 50.0 g C) 112g 6. When solutions of cobalt(l) chloride and carbonic acid react, which of the following terms will be present in the balanced net ionic equation? A) CoCO3s) B) H2 2(ag) C) 2CoCOss) D) 2C1 (aa) E) two of these 7. Sullarmic acid, HSO:NH2 (MM = 97.11 g/mol) is a monoprotic acid that can be used to standardize(titrate) a strong base: A0.179 g sample of HSO&NH2; required 19.4 ml of an aqueous solution of KOH for complete reaction. What is the molarity of the KOH solution? A) 9.25 M B) 9.50 x 105 M C) 0.0950 M D) 0.194 M E) none of these

Explanation / Answer

5)

Molar mass of Ca3(PO4)2 = 3*MM(Ca) + 2*MM(P) + 8*MM(O)

= 3*40.08 + 2*30.97 + 8*16.0

= 310.18 g/mol

mass of Ca3(PO4)2 = 103.0 g

we have below equation to be used:

number of mol of Ca3(PO4)2,

n = mass of Ca3(PO4)2/molar mass of Ca3(PO4)2

=(103.0 g)/(310.18 g/mol)

= 0.3321 mol

Molar mass of H2SO4 = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass of H2SO4 = 75.0 g

we have below equation to be used:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(75.0 g)/(98.086 g/mol)

= 0.7646 mol

we have the Balanced chemical equation as:

2 Ca3(PO4)2 + 3 H2SO4 ---> 2 H3PO4 + 3 CaSO4

2 mol of Ca3(PO4)2 reacts with 3 mol of H2SO4

for 0.3321 mol of Ca3(PO4)2, 0.4981 mol of H2SO4 is required

But we have 0.7646 mol of H2SO4

so, Ca3(PO4)2 is limiting reagent

we will use Ca3(PO4)2 in further calculation

Molar mass of H3PO4 = 3*MM(H) + 1*MM(P) + 4*MM(O)

= 3*1.008 + 1*30.97 + 4*16.0

= 97.994 g/mol

From balanced chemical reaction, we see that

when 2 mol of Ca3(PO4)2 reacts, 2 mol of H3PO4 is formed

mol of H3PO4 formed = (2/2)* moles of Ca3(PO4)2

= (2/2)*0.3321

= 0.3321 mol

we have below equation to be used:

mass of H3PO4 = number of mol * molar mass

= 0.3321*97.99

= 32.5 g

Answer: 32.5 g

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