stoichiometry

Moles

100

How many moles of oxygen atoms are in 0.5 moles of H₂O₂?

1 mole of oxygen atoms

2*0.5=1

200

How many atoms are in 0.1 mole of CH₃OH?

3.6*10²³ atoms

6*0.1*6*10²³=3.6*10²³ 

300

How many moles of Cl^- are in 2 mole of molten MgCl₂?

4 moles of Cl^-

400

How many moles of ions are in 0.2 moles of Al₂(SO₄)₃?

0.2*5=1 mole of ions

2 Al₃⁺

3 SO₄^2-

500

How many boxes of eggs (each containing a dozen) can we make out of 1 mole of eggs?

6*10²³/12=0.5*10²³=5*10²² boxes

Masses

100

What is the mass of 0.1 mole of C₆H₁₂O₆?

180*0.1=18g

200

How many moles are in 49 g of H₂SO₄?

0.5 mole H₂SO₄

300

The mass of 0.2 moles of a substance is 8.8g. What is the molar mass of the substance?

44 grams

400

A sample of 7g consists of 0.25 moles of diatomic molecules of a certain element. What is the molecular mass of the element?

14g (N)

since the molecules are diatomic, they consist of 0.5 moles of the element

0.5 mole------7g

1 mole-----Xg    >>7/0.5=14

500

The mass of 1 mole of X₂O₃ is 160 g. what is the molar mass of X?

56 grams

(160-3*16)/2=56

Empiric formula

100

What is the molecular formula of a compound that consists of only carbon and hydrogen in a ratio of 1:1, and has a molar mass of 78 g/mol?

78/(1+12)=6

C₆H₆

200

In a sample are found 1.33 moles of N (nitrogen) and 2 moles of O (oxygen). what is the empirical formula?

N₂O₃

300

Analysis of 15 grams of a compound shows that it consists of 6g carbon, 1g hydrogen and 8g oxygen. What is the empirical formula?

CH₂O

n(C)=6/12=0.5        /*2

n(H)=1/1=1            /*2

n(O)=8/16=0.5        /*2

400

The empirical formula consists of elements X and Y.

A_r(X)=20

A_r(Y)=15

It was found that the sample consists of 40% element X. what is the empirical formula?

XY₂

n(X)=40/20=2    /:2

n(Y)=60/15=4     /:2

500

Combustion of 2.4g Mg results with 4.0g of compound. what is the empirical formula of the compound?

MgO

2.4g Mg -->0.1 mole Mg

4-2.4=1.6g O -->0.1 mole O

Reaction calculations

100

In combustion of ethanol (C₂H₅OH) the products are carbon dioxide and water. Formulate and balance the equation.

C₂H₅OH + 3O₂ --> 2CO₂ + 3H₂O

200

How much CO₂ (in mass) would be produced from combustion of 11.5 g ethanol?

22g of CO₂

C₂H₅OH + 3O₂ --> 2CO₂ + 3H₂O

1/4 mole of ethanol --> 1/2 mole of CO₂=>22g

300

How much oxygen (in mass) is required to burn 460g of ethanol?

960g of O₂

C₂H₅OH + 3O₂ --> 2CO₂ + 3H₂O

460g=10 mole of ethanol=> need 30 moles of O₂

the mass of each mole of O₂ is 32g, so we need 30*32=960g of O₂

400

Determine the mass of ethanol that was burned when 540g of water were produced.

460g ethanol was burned

C₂H₅OH + 3O₂ --> 2CO₂ + 3H₂O

46g ethanol ----18*3=54g water

500

In room of size 3m*3m*3m there are about 60 moles of oxygen (O₂).

If we try to burn 1Kg ethanol, what will finish up first - the ethanol or the oxygen? (must explain)

The oxygen will finish up first.

60 moles of O₂ will burn 20 moles of ethanol which are 20*46=920g ethanol, but we have 1000g ethanol, we have excess of ethanol.

C₂H₅OH + 3O₂ --> 2CO₂ + 3H₂O