Question

I need to see how to do 2 and 3

Answer 1

Answer:

Explanation:

a) Here, we want to get the limiting reactant

The limiting reactant forms less of the product

From the equation of reaction:

4 moles of aluminum gave 2 moles of product

0.32 mol will give x moles of product:

4 * x = 0.32 * 2

x = 0.16 moles

3 moles of oxygen gave 2 moles of product

0.26 mol will give x moles of product

3 * x = 0.26 * 2

x = 0.173 moles

That means that aluminum is the limiting reagent

b) From the question:

6.38 * 10^-3 mol oxygen gives x mol of product

3 moles of oxygen gives 2 moles of product

To get x:

$x\text{ = }\frac{6.38\times10^{-3}\times2}{3}\text{ = }0.00213\text{ mole}$

For aluminium, we do same process:

$x\text{ = }\frac{9.15\times10^{-3}\times2}{4}\text{ = 0.004575 mole}$

c) From the mass available, we need to get the number of moles that could be produced from each

To get the number of moles, we have to divide the mass by the atomic mass of the element

For Aluminium, the atomic mass is 27 amu

That means the number of moles is:

$\frac{3.17}{27}\text{ = 0.1174 mole}$

Now, from the equation of reaction:

4 moles aluminum gave 2 moles oxygen

0.1174 mole aluminum will give:

$\frac{0.1174\times2}{4}\text{ = 0.0587 mole}$

For oxygen, the atomic mass is 16 amu

For molecular oxygen, we have the molar mass as 32g/mol

The number of moles that will react is thus:

$\frac{2.55}{32}\text{ = 0.0796875 mole}$

From the equation of reaction:

3 moles of oxygen gave 2 moles of product,

0.159375 mole will give:

$\frac{0.0796875\times2}{3}\text{ = 0.053125 mole}$

Since the number of moles from oxygen is lesser, it translates to a lesser amount of product and that makes oxygen the limiting reactant in this case