Question

object the following




askedto the following​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Fe_2O_3

= 0.2128 x

Weight of the rust,

\frac{0.2128 x}{159.7 g/mol}

Moles of rust =

1 mole of rust has 2 moles of iron.

2\times \frac{0.2128 x}{159.7 g/mol}

Moles of iron in rust =

2\times \frac{0.2128 x}{159.7 g/mol}\times 55.85 g/mol=0.1488 x

Weight of moles of iron:

Weight of iron calculated above is the iron which has been present in the form of rust on iron.

Percent of iron that has been rusted:

\frac{0.1488 x}{x}\times 100=14.88\%

14.88% of iron has been rusted