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.

where is the following

explanation:

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.