6. Find the number of atoms in 3.8mg of NaPO? 5 pts 7. A 95.0-g sample of Fe0s r

Question

6. Find the number of atoms in 3.8mg of NaPO? 5 pts 7. A 95.0-g sample of Fe0s reacts with Oz to produce FeOs: Determine the number of g of Fe Os produced. b. If the % yield of Fed) in Problem 7 is 25.0%, what is the actual yield of FeO? 5 pts 8. Give the proper name for the following compounds. 12 pts a. Mg(OH)2 b. Co(CN) d. Cu(SO) 9. Write the correct formula for the following compounds: 12 pts a. Calcium bromide CaBr2 Manganese(lI) Chlorate(C103) c. DiBoron Trioxide D203 unafa 2(S04)3 f. Phosphorus Trihydride Ptta 10. According to the VSEPR theory, what is the geometry (shape) of the following Lewis structure? 12 pts 11. Predict the correct formula of the following compound using the Lewis Dot structure. 12 pts c. N Br d. s +Se a. Al +P 13. Calculate the number of gof CO, produced from the combustion of 11.0 mol of C Hs using the balance equation: CHS (g) + 502(g) 3CO2(g) + 4H:00g).

Explanation / Answer

6. Find the number of atoms in 3.8 mg of Na3PO4?

Given, weight of Na3PO4 = 3.8 mg = 0.0038 g

no. of moles = wt/mol.wt = 0.0038g/163.94 g/mol = 2.318x10-5 moles

Therefore no. of moles of Na atoms = 3 x 2.318x10-5 moles

no. of moles of P atoms = 2.318x10-5 moles

no. of moles of O atoms = 4x2.318x10-5 moles

Hence, no. of Na atoms = 3 x 2.318x10-5 mole x 6.022 × 1023 atoms/mole = 4.188 x 1019 atoms

no. of P atoms = 2.318x10-5 mole x 6.022 × 1023 atoms/mole = 1.4 x 1019 atoms

no. of O atoms = 4 x 2.318x10-5 mole x 6.022 × 1023 atoms/mole = 5.58 x 1019 atoms

  

Therefore, total no. of atoms = no. of Na atoms + no. of P atoms + no. of O atoms

= 4.188 x 1019 + 1.4 x 1019 + 5.58 x 1019

= 1.17 x 1020 atoms

Hence, the asnwer is = 1.17 x 1020 atoms

7. A 95.0 g of the sample of Fe3O4 reacts with O2 to produce Fe2O3;

4Fe3O4(s) + O2 (g) ----> 6Fe2O3(s)

The no. of g of Fe2O3 produced?

4 moles of Fe3O4 reacts and produces 6 moles of Fe2O3

no. of moles of Fe3O4 reacted = wt/mol.wt = 95 g/ 231.5326 g/mol = 0.41 mol

no. of moles of Fe2O3 produced = (6/4)x 0.41 mol = 0.6154 moles

the mass of Fe2O3 produced = no. moles x mol.wt = 0.6154 moles x 159.69 g/mol = 98.28 g

Hence, the mass of Fe2O3 produced = 98.28 g

b. percent yield = (actual yield/theoretical yield) x 100%

Percent yield = 25%

Actual yield =?

Theoretical yield = 98.28 g

Therefore, 25% = (actual yield/ 98.28)x100%

actual yield = 24.57 g

Hence, the actual yield = 24.57 g

8.

a. Mg(OH)2 - Magnesium hydroxide

b. Co(CN)2 - Cobalt(II) cyanide

d. Cu2(SO3)3 - copper sulfite

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