Question 1

IN 2 H2 + O2 → 2 H2O, HOW MANY MOLES OF WATER FORM FROM 6 MOLES OF H2 WITH EXCESS OXYGEN?

Accepted Answer

6 MOLES. Hydrogen and water appear in a 2 : 2 ratio, which simplifies to 1 : 1, so 6 mol of hydrogen gives 6 mol of water.

Question 2

IN N2 + 3 H2 → 2 NH3, HOW MANY MOLES OF H2 ARE NEEDED TO MAKE 10 MOLES OF NH3?

Accepted Answer

15 MOLES. Hydrogen and ammonia react in a 3 : 2 ratio, so 10 mol of ammonia needs 10 × 3/2 = 15 mol of hydrogen.

Question 3

CALCIUM CARBONATE, CaCO3, HAS A MOLAR MASS OF 100 g/mol. WHAT IS THE MASS OF 0.25 mol OF IT?

Accepted Answer

25 g. Mass equals moles times molar mass: 0.25 mol × 100 g/mol = 25 g. Dividing instead of multiplying gives the 400 g trap.

Question 4

IN 2 H2 + O2 → 2 H2O, 4 mol OF H2 IS MIXED WITH 3 mol OF O2. WHICH IS THE LIMITING REACTANT?

Accepted Answer

H2. Four moles of hydrogen need only 2 mol of oxygen. There are 3 mol of oxygen available, so hydrogen runs out first and sets the amount of product.

Question 5

A REACTION HAS A THEORETICAL YIELD OF 40 g. YOU COLLECT 30 g OF DRY PRODUCT. WHAT IS THE PERCENT YIELD?

Accepted Answer

75%. Percent yield is actual over theoretical times 100: 30 ÷ 40 × 100 = 75%. Putting the numbers the other way round gives the impossible 133%.

Question 6

A STUDENT REPORTS A PERCENT YIELD OF 112%. WHAT IS THE MOST LIKELY CAUSE?

Accepted Answer

THE PRODUCT WAS STILL WET WHEN WEIGHED. Every other option lowers the yield. A yield above 100% means the weighed product contains something extra, most often water or solvent that was not dried off.

Question 7

A COMPOUND IS 40.0% CARBON, 6.7% HYDROGEN AND 53.3% OXYGEN BY MASS. WHAT IS ITS EMPIRICAL FORMULA?

Accepted Answer

CH2O. In 100 g there are 40.0 ÷ 12 = 3.33 mol of carbon, 6.7 ÷ 1 = 6.7 mol of hydrogen and 53.3 ÷ 16 = 3.33 mol of oxygen. Dividing by 3.33 gives 1 : 2 : 1, so CH2O.

Question 8

A COMPOUND HAS THE EMPIRICAL FORMULA CH2O AND A MOLAR MASS OF 180 g/mol. WHAT IS ITS MOLECULAR FORMULA?

Accepted Answer

C6H12O6. One CH2O unit has a mass of 12 + 2 + 16 = 30 g/mol. 180 ÷ 30 = 6, so every subscript is multiplied by 6 to give C6H12O6, which is glucose.

Question 9

WHEN C3H8 + O2 → CO2 + H2O IS BALANCED WITH THE SMALLEST WHOLE NUMBERS, WHAT IS THE COEFFICIENT OF O2?

Accepted Answer

5. Balance carbon first (3 CO2), then hydrogen (4 waters). That puts 6 + 4 = 10 oxygen atoms on the right, which takes 5 oxygen molecules.

Question 10

THE COEFFICIENTS IN A BALANCED EQUATION COMPARE AMOUNTS IN MOLES, NEVER IN ___.

Accepted Answer

GRAMS. Two moles of hydrogen react with one mole of oxygen, but by mass that is about 4 g of hydrogen to 32 g of oxygen. Convert masses to moles before using the ratio.

Question 11

GLUCOSE, C6H12O6, AND FORMALDEHYDE, CH2O, HAVE DIFFERENT MOLECULAR FORMULAS BUT THE SAME ___ FORMULA.

Accepted Answer

EMPIRICAL. Dividing the subscripts of C6H12O6 by 6 gives CH2O, so both compounds reduce to the same simplest whole-number ratio of atoms.

Stoichiometry