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

2Ca(s) + O2(g)--> 2CaO(s)
Calculate the mass of calcium prepared from 4,20g of Ca and 2.80g of O2. (a) How many moles of CaO can form from the given mass of Ca? (b) How many moles of CaO can form from the given mass of O2? (c)What is the limiting reactant? (d) How many grams of CaO can be formed?

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To answer these questions, we need to use the concept of stoichiometry. Stoichiometry is a relationship between the relative quantities of substances taking part in a reaction.

First, we'll calculate the number of moles of calcium (Ca) and oxygen gas (O2) based on their given masses using their respective molar masses.

(a) How many moles of CaO can form from the given mass of Ca?
Given: Mass of Ca = 4.20g
Molar mass of Ca = 40.08 g/mol

Using the formula: Moles = Mass / Molar Mass
Moles of Ca = 4.20g / 40.08 g/mol ≈ 0.1047 mol

Since the balanced equation tells us that 2 moles of Ca correspond to 2 moles of CaO, we can say that from 0.1047 mol of Ca, 0.1047 mol of CaO can form.

(b) How many moles of CaO can form from the given mass of O2?
Given: Mass of O2 = 2.80g
Molar mass of O2 = 32.00 g/mol

Using the formula: Moles = Mass / Molar Mass
Moles of O2 = 2.80g / 32.00 g/mol ≈ 0.0875 mol

Since the balanced equation tells us that 1 mole of O2 corresponds to 2 moles of CaO, we can say that from 0.0875 mol of O2, 0.0875 x 2 = 0.175 mol of CaO can form.

(c) What is the limiting reactant?
The limiting reactant is the reactant that is completely consumed first, thereby limiting the amount of product that can be formed.

To determine the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
From part (a), we have 0.1047 mol of CaO possible from Ca.
From part (b), we have 0.175 mol of CaO possible from O2.

We can see that the amount of CaO formed from Ca is less than the amount formed from O2. Therefore, Ca is the limiting reactant.

(d) How many grams of CaO can be formed?
To find the mass of CaO formed, we use the stoichiometric ratio from the balanced equation.

From the balanced equation: 2Ca(s) + O2(g) --> 2CaO(s)
We see that for every 2 moles of CaO, we need 2 moles of Ca.

Since we have established that Ca is the limiting reactant, we will base the calculation on the moles of Ca.

Using the formula: Mass = Moles x Molar Mass
Molar mass of CaO = 56.08 g/mol

Mass of CaO = 0.1047 mol x 56.08 g/mol ≈ 5.88 g

Therefore, approximately 5.88 grams of CaO can be formed.