How many grams of iron can be produced from 3.65g of Iron (III) Sulfate? 3Mg + Fe2(SO4)3 --> 3MgSO4 + 2Fe A. 1.02g Fe B. 5.41g Fe C. 0.548g Fe D. 7.56g Fe

Option A. 1.02g of Fe.

Explanation:

Step 1:

The balanced equation for the reaction.

3Mg + Fe2(SO4)3 —> 3MgSO4 + 2Fe

Step 2:

Determination of the mass Fe2(SO4)3 that reacted and the mass of Fe produced from the balanced equation. This is illustrated below:

Molar Mass of of Fe2(SO4)3 = (2x56) + 3[32 + (16x4)] = 112 + 3[32 + 64] = 112 + 3[96] = 400g/mol

Mass of Fe2(SO4)3 from the balanced equation = 1 x 400 = 400g

Molar Mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 2 x 56 = 112g

From the balanced equation above,

400g of Fe2(SO4)3 reacted to produce 112g Fe.

Step 3:

Determination of the mass of Fe produced by reacting 3.65g of Fe2(SO4)3.

This is illustrated below:

From the balanced equation above,

400g of Fe2(SO4)3 reacted to produce 112g Fe.

Therefore, 3.65g of Fe2(SO4)3 will react to produce = (3.65 x 112)/400 = 1.02g of Fe.

Therefore, 1.02g of Fe is produced from 3.65g of Fe2(SO4)3.

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

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