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 7.97 g of Ca and 5.31 g of O2.

Answer 1

11.11 g

Step-by-step explanation:

The first thing is to calculate the number of moles of each of the reactants so that we establish which one is the limited reactant because that is the one we use for our calculations.

number of moles of Ca = mass/molar mass = {7.97 g/40.078 g/mol = 0.19811 mol

number of moles of Oxygen = 5.31 g/31.988 g/mol = 0.1660 mol

for all the oxygen to be used up it would require 0.332 mol of Ca (Since their stoichiometric ratio is 1:2). So Ca is a limiting reactant and Oxygen is in excess. So we use Ca's number of moles for calculation moving forward.

mass of CaO produced = mol * molar mass = 0.19811 mol * 56.069 g/mol

= 11.11 g

Answer 2

11.14 grams of CaO

Step-by-step explanation:

The reaction is:

`2Ca(s) + O_(2)(g) -->2CaO(s)`

As per equation one mole of oxygen gas will reaction with two moles of calcium to give two moles of calcium oxide.

the moles of oxygen reacted =
`(mass)/(molarmass)=(5.31)/(32)= 0.166`

moles of calcium reacted =
`(mass)/(atomicmass)=(7.97)/(40)=0.199`

the moles of oxygen that will react with 0.199 moles of Ca =
`(0.199)/(2)= 0.0995`

The moles of CaO produced = 0.199

Mass of calcium oxide produced = moles X molar mass

= 0.199 X 56 = 11.14g

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