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The Mole Concept for Compounds Practice Problems 1. Determine the number of C&H1206 atoms in a 5.86 g sample of CH1206 2. Determine the number of moles of oxygen in a 4.88 mole sample of H2COs. 3. Determine the number of grams of carbon in an 8.63 g sample of C3HsCIO. 4. Determine the number of oxygen atoms in 64.85 g sample of aluminum sulfate.

Answer 1

Ans 1

Moles of C6H12O6 = mass/molecular weight

= 5.86g / 180.156 g/mol

= 0.032527 mol

Number of atoms = moles x Avogadro number

= 0.032527 mol x 6.023 x 10^23 atoms/mol

= 1.959 x 10^22 atoms

Ans 2

Moles of H2CO3 = 4.88 mol

Moles of O = Moles of H2CO3 x 3 mol O / 1 mol H2CO3

= 4.88 x 3 / 1

= 14.64 mol

Ans 3

Mass of C3H5ClO = 8.63 g

Mass of C

= Mass of C3H5ClO x (3*12 g C / 92.52 g C3H5ClO)

= 8.63 x 36 / 92.52

= 3.357 g

Ans 4

Mass of Aluminium sulfate Al2(SO4)3 = 64.85 g

Moles of Al2(SO4)3 = mass/molecular weight

= 64.85g / 342.15g/mol

= 0.1895 mol

Moles of O

= Moles of Al2(SO4)3 x 12 mol O / 1 mol Al2(SO4)3

= 0.1895 mol x 12/1

= 7.96 mol

Number of O atoms = moles of O x Avogadro number

= 7.96 mol x 6.023 x 10^23 atoms/mol

= 4.794 x 10^24 atoms