This tutor shows how a balanced equation links amounts of reactants and products. You type the equation, give the amount of each reactant, and the tool walks through the mole reasoning in four steps, with units. It works in an ideal model: the reaction goes to completion and there are no side reactions.
It supports the ideas taught in using an equation ratio to relate reacting amounts.
How do I use it?
- Choose an example equation from the menu, or type your own in the equation box. Use
->between the two sides and+between species. - Enter the amount of each reactant. Each has its own box and a unit menu, mol or g. The relative formula mass is worked out from the formula.
- Leave percentage yield at 100 for the ideal model, or enter a lower value to see a scaled product mass.
- Press “Work it out.” If the equation is not balanced, a message lists the elements that do not match. Balance it and try again.
- Read the four steps in the result, then change an amount and run it again to see what shifts.
How do I read the result?
Step 1 converts each amount to moles, using moles = mass / Mr where the amount was given in grams. Step 2 divides each mole value by its coefficient from the equation. This number is the mole ratio check.
Step 3 finds the limiting reactant: the one with the smallest value after dividing by its coefficient. It runs out first and sets how much product forms. If two or more share the smallest value, the reactants are in exactly the stoichiometric ratio and none is left over. Any other reactant is in excess, and the tool shows how many moles and grams remain.
Step 4 multiplies the coefficient of each product by that smallest value to get moles, then by Mr to get grams. The last column shows the mass at your stated yield.
Example walk-through
First example. Select 2H2 + O2 -> 2H2O. The tool starts with 2 mol of H2 and 1 mol of O2.
Step 2 gives 2 / 2 = 1 for H2 and 1 / 1 = 1 for O2. Both values are equal, so the reactants are in exactly the stoichiometric ratio and none is left over.
For H2O, 2 x 1 = 2 mol, and with Mr = 18 that is 2 x 18 = 36 g at 100 per cent yield. This matches the ideal model: 2 mol H2 and 1 mol O2 form 2 mol H2O.
Second example. Select Mg + 2HCl -> MgCl2 + H2. Enter 4.8 g for Mg and 0.2 mol for HCl.
Mg has Mr = 24, so 4.8 / 24 = 0.2 mol. Dividing by coefficients gives 0.2 / 1 = 0.2 for Mg and 0.2 / 2 = 0.1 for HCl.
HCl is the smaller, so HCl is the limiting reactant. Mg is in excess: 0.2 - 0.1 = 0.1 mol, which is 2.4 g.
Products use the smaller value, 0.1. MgCl2 is 0.1 mol, and with Mr = 24 + 2 x 35.5 = 95 that is 9.5 g.
H2 is 0.1 mol, which is 0.2 g. At an 80 per cent yield the last column shows 7.6 g of MgCl2 and 0.16 g of H2.
The trap here is using the amounts directly without dividing by the coefficients. Raw moles look equal, 0.2 mol of each, so the 1 : 2 ratio is ignored and both seem to run out together. The ratio step is where the reasoning belongs.
Assumptions and limits
- The model is ideal: complete reaction, no side reactions, and products counted from the limiting reactant.
- Only a fixed list of common elements is included. An unknown symbol gives a message.
- It checks that the equation balances but does not balance it for you.
- It works with moles and masses on the page. Gas volumes and solution concentrations are not included here.
- It is an educational model only, not a procedure for making or handling any substance.
- Check your own syllabus for which calculations and relative atomic masses you must use.
Which lessons explain the output?
Start with calculating a relative formula mass, then converting mass to amount using consistent units and using an equation ratio to relate reacting amounts. Step 4 is checked in checking an amount calculation by units. The whole topic is on the relative masses and amounts module and tested in the relative masses and amounts practice.
For the common error this tool exposes, see why correct arithmetic can use the wrong chemical ratio, and the mole ratio clinic. Balancing comes first in the equation balance reasoning trainer. The tools page lists the rest.
Mole questions often go wrong at one specific step, which a teacher can spot quickly by reading your working. That is the kind of help available in online one-to-one Chemistry tuition.