Question

Calculate how many grams of Aluminum are needed to produce 21.6 grams of Aluminum oxide (Al2O3).

4Al + 3O2 → 2Al2O3

a
11.43 g Al
b
26.982 g Al
c
5.71 g Al
d
132 g Al

Answer 1

Answer: A. 11.43 g Al

Step-by-step explanation:

To solve this problem, we need to use the process of dimensional analysis, which essentially turns all of the values into fractions that will cancel out units. We always start by using what we are given and then set up the units so that everything cancels besides what we need to solve for.

given Al2O3 g * (1molAl2O3/mass Al2O3 for periodic table) * (Al mole ratio from balanced equation / Al2O3 mole ratio from the balanced equation) * ( Al mass from periodic table / 1 mol Al)

Now, I will plug all values in and solve.

21.6 gAl2O3 * (1molAl2O3/ 101.96 g Al2O3) * (4 mol Al / 2 mol Al2O3) * (26.98 g Al / 1 mol Al)

Now, multiply across the numerator and denominator separately and then proceed with division.

11.43 g Al

Hope this helps!

Answer 2

1.Write down the molecular formula Al2O3

2. Note down the atomic masses of Aluminum & oxygen. ( 23 & 16 )

3.Find its molecular mass ( 2 × 23 ) + ( 3 × 16 ) units = 46 + 48 = 94 units

4.Find mass of Aluminum in 1 mole of aluminum oxide ,46 grams in 94 grams

5. Find mass of aluminum in the given mass of aluminum oxide

94 grams of aluminum oxide contains 46 grams of aluminum.

21.6 grams of aluminium oxide will contain ( 46 / 94 ) × 21.6 grams of aluminum.

Do the calculation

Hopefully this sep-by-step equation will help, if the answer isn’t right then maybe try it to the nearest one that is listed as your answer option. Have a good day!