Question

a piece of iron when kept in air increases its weight by 21.28% what percent of iron has been rusted? ​

Answer 1

Answer:

14.88% of iron has been rusted.

Explanation:

Let the weight of u rusted iron be x

Weight of the iron after rusting = 0.2128 x + x= 1.2128 x

Weight gained by the iron = 21.28% of x = 0.2128 x

Weight of the rust,$Fe_2O_3$ = 0.2128 x

Moles of rust = $\frac{0.2128 x}{159.7 g/mol}$

1 mole of rust has 2 moles of iron.

Moles of iron in rust =$2\times \frac{0.2128 x}{159.7 g/mol}$

Weight of moles of iron: $2\times \frac{0.2128 x}{159.7 g/mol}\times 55.85 g/mol=0.1488 x$

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.

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

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