VDOE SOL (multiple choice)-style practice: For N_2 + 3\,H_2 → 2\,NH_3, how many moles of ammonia form from 6.0 mol of hydrogen? (A) 2.0 mol (B) 4.0 mol (C) 6.0 mol (D) 9.0 mol

The answer is (B) 4.0 mol. The mole ratio of hydrogen to ammonia from the balanced coefficients is 3 : 2. So 6.0 mol of hydrogen gives 6.0 × 23 = 4.0 mol of ammonia. The trap is using a 1 : 1 ratio; always read the ratio from the balanced coefficients, here 3 hydrogen to 2 ammonia.

VDOE SOL (tech-enhanced, fill in the blank)-style practice: Given 2\,Mg + O_2 → 2\,MgO, calculate the mass of magnesium oxide formed from 4.86 g of magnesium. (Molar masses: Mg 24.3, MgO 40.3 g/mol.)

A 3-point mass-to-mass item. Step 1, mass to moles: moles Mg = 4.8624.3 = 0.200 mol (1 point). Step 2, mole ratio: Mg to MgO is 2 : 2, so 0.200 mol Mg gives 0.200 mol MgO (1 point). Step 3, moles to mass: mass MgO = 0.200 × 40.3 = 8.06 g (1 point). Markers reward the full chain: grams to moles, mole ratio, moles to grams, with the magnesium oxide mass exceeding the magnesium mass because oxygen has combined.

VirginiaChemistrySyllabus dot point

How does a balanced equation let you calculate the amount of product from the amount of reactant?

Stoichiometry and the mole ratio: use the mole ratio from a balanced equation to convert between moles and masses of reactants and products, including gas volumes at STP.

A focused Virginia SOL Chemistry answer on stoichiometry under CH.3: reading the mole ratio from a balanced equation, mole-to-mole and mass-to-mass calculations, and using the molar volume of a gas at STP, with the full three-step chain.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The mole ratio
Mole-to-mole calculations
Mass-to-mass calculations
Gas volumes at STP
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What this topic is asking

Standard CH.3 brings the mole and balanced equations together in stoichiometry. Virginia expects you to read the mole ratio from a balanced equation and use it to convert between moles and masses of reactants and products, and to use the molar volume of a gas at standard temperature and pressure (STP). This is the quantitative heart of the course and a frequent constructed-response task.

The mole ratio

The coefficients count particles in proportion, so they are read as moles. A balanced equation must come first, because an unbalanced equation gives the wrong ratio. To use the ratio, multiply the known amount by the fraction with the wanted substance on top.

Mole-to-mole calculations

The simplest stoichiometry converts moles of one substance directly to moles of another using the ratio. If $3$ mol of hydrogen reacts in $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$ , then the hydrogen-to-ammonia ratio of $3 : 2$ gives $3 \times \dfrac{2}{3} = 2$ mol of ammonia. No mass conversion is needed when the question is in moles on both ends.

Mass-to-mass calculations

When the question gives a mass and asks for a mass, use a three-step chain:

Step	Operation
1	Convert the given mass to moles: divide by its molar mass
2	Apply the mole ratio from the balanced coefficients
3	Convert the target moles to mass: multiply by its molar mass

This "grams to moles, mole ratio, moles to grams" chain works for any reactant-to-product or reactant-to-reactant pairing. Set the calculation out with units so they cancel.

Gas volumes at STP

So a gas-producing reaction can be taken one step further: find moles of gas by stoichiometry, then multiply by $22.4$ L/mol to get the volume at STP. The $22.4$ L value applies only at STP.

A full mass-to-mass problem

Given $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$ , calculate the mass of water formed from $8.0$ g of oxygen. (Molar masses: $\text{O}_2$ $32.0$ , $\text{H}_2\text{O}$ $18.0$ g/mol.)

step 1 Mass of oxygen to moles

$\text{moles O}_2 = \dfrac{8.0\ \text{g}}{32.0\ \text{g/mol}} = 0.25$ mol.

step 2 Apply the mole ratio

The ratio $\text{O}_2 : \text{H}_2\text{O}$ is $1 : 2$ , so $0.25$ mol O $_2$ gives $0.50$ mol water.

step 3 Moles of water to mass

$\text{mass H}_2\text{O} = 0.50\ \text{mol} \times 18.0\ \text{g/mol} = 9.0$ g.

step 4 Sense-check

The water mass ( $9.0$ g) is greater than the oxygen mass because the hydrogen that combined adds mass, consistent with conservation of mass.

Try this

Q1. For $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ , how many moles of carbon dioxide form from $3.0$ mol of carbon? [1 point]

Cue. The ratio is $1 : 1$ , so $3.0$ mol of carbon gives $3.0$ mol of carbon dioxide.

Q2. What volume does $2.0$ mol of oxygen gas occupy at STP? [1 point]

Cue. $2.0 \times 22.4 = 44.8$ L.

Exam-style practice questions

Practice questions written in the style of VDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SOL (multiple choice)1 marksFor

\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3

, how many moles of ammonia form from

6.0

mol of hydrogen? (A)

2.0

mol (B)

4.0

mol (C)

6.0

mol (D)

9.0

mol

Show worked answer →

The answer is (B) $4.0$ mol.

The mole ratio of hydrogen to ammonia from the balanced coefficients is $3 : 2$ . So $6.0$ mol of hydrogen gives $6.0 \times \dfrac{2}{3} = 4.0$ mol of ammonia.

The trap is using a $1 : 1$ ratio; always read the ratio from the balanced coefficients, here $3$ hydrogen to $2$ ammonia.

SOL (tech-enhanced, fill in the blank)3 marksGiven

2\,\text{Mg} + \text{O}_2 \rightarrow 2\,\text{MgO}

, calculate the mass of magnesium oxide formed from

4.86

g of magnesium. (Molar masses: Mg

24.3

, MgO

40.3

g/mol.)

Show worked answer →

A 3-point mass-to-mass item.

Step 1, mass to moles: $\text{moles Mg} = \dfrac{4.86}{24.3} = 0.200$ mol (1 point).
Step 2, mole ratio: Mg to MgO is $2 : 2$ , so $0.200$ mol Mg gives $0.200$ mol MgO (1 point).
Step 3, moles to mass: $\text{mass MgO} = 0.200 \times 40.3 = 8.06$ g (1 point).

Markers reward the full chain: grams to moles, mole ratio, moles to grams, with the magnesium oxide mass exceeding the magnesium mass because oxygen has combined.

Related dot points

Sources & how we know this

2018 Science Standards of Learning - Chemistry — Virginia Department of Education (2018)
Chemistry Curriculum Framework — Virginia Department of Education (2018)