Question

Many metals react with oxygen gas to form the metal oxide. For example, calcium reacts as follows:

2Ca(s) + O2(g) → 2CaO(s)

Calculate the mass of calcium oxide that can be prepared from 6.87 g of Ca and 4.58 g of O2.

***The Answer is not 16.03g -- I have already tried it but the program does not accept that answer***

Answer 1

Molar mass of Ca = 40.08 g/mol

mass(Ca)= 6.87 g

number of mol of Ca,

n = mass of Ca/molar mass of Ca

=(6.87 g)/(40.08 g/mol)

= 0.1714 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 4.58 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(4.58 g)/(32 g/mol)

= 0.1431 mol

Balanced chemical equation is:

2 Ca + O2 ---> 2 CaO +

2 mol of Ca reacts with 1 mol of O2

for 0.171407 mol of Ca, 0.085704 mol of O2 is required

But we have 0.143125 mol of O2

so, Ca is limiting reagent

we will use Ca in further calculation

Molar mass of CaO,

MM = 1*MM(Ca) + 1*MM(O)

= 1*40.08 + 1*16.0

= 56.08 g/mol

According to balanced equation

mol of CaO formed = (2/2)* moles of Ca

= (2/2)*0.171407

= 0.171407 mol

mass of CaO = number of mol * molar mass

= 0.1714*56.08

= 9.61 g

Answer: 9.61 g