Question

How many grams of a metal are needed to react with oxygen if 46 g of rust is to be formed?

Answer 1

`32.20\text{ g }`

Step-by-step explanation:

The first thing we have to write here is the equation of reaction that forms rust

We have that as:

`4Fe_((s))\text{ + 3O}_(2(g))\text{ }\rightarrow\text{ 2Fe}_2O_(3(s))`

From the equation of reaction, 4 moles of iron react with 3 moles of oxygen to give 2 moles of rust

Now, let us get the actual number of moles

To get the number of moles of rust formed, we have to divide the mass of rust formed by the molar mass of rust

The molar mass of rust is 160 g/mol

The number of moles of rust formed is thus:

`\frac{46\text{ g}}{160\text{ g/mol}}\text{ = }(23)/(80)\text{ mol}`

From the equation of reaction:

4 moles of iron give 2 moles of rust

x moles of iron will give 23/80 moles of rust

Mathematically:

`\begin{gathered} 4\text{ }*(23)/(80)\text{ = 2}*\text{ x} \\ \\ x\text{ = 4 }*(1)/(2)*(23)/(80)\text{ = }(46)/(80)\text{ mol} \end{gathered}`

Now, to get the mass of the metal reacted, we have to multiply the atomic mass of iron by the number of moles above

The atomic mass of iron is 56 amu

Thus, we have the mass of the metal formed as:

`56\text{ }*(46)/(80)\text{ = 32.2 g}`