Extra Stoichiometry Practice
1. How many grams of aluminum oxide can be formed by the reaction of 38.8 g of aluminum with
oxygen?
4 Al (8) + 3 O2(g) →2 A12O(s)

50. 4g

Explanation:

First calculate number of moles of aluminium in 38.8g

Moles = 38.8g/ 26.982mol/g

= 1.44mol

By looking at the balance equation you can see that 4 moles of aluminium produce 2 moles of aluminium oxide.

4 = 2

1.4 = x

Find the value of x

x= (1.4×2)/4= 0.72 mol

0.72 moles of aluminium oxide are produced from 38.8g of aluminium

Now find the mass of aluminium produced.

Mass = moles × molar mass

= 0.72mol × 69.93 mol/g

= 50.4g

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.