How To Do Stoichiometry With Pictures

Don’t forget to multiply through by a coefficient or subscript if one is present. For example, H2SO4 + Fe —> Fe2(SO4)3 + H2 On the reactant (left) side of the equation there are 2 H atoms (H2), 1 S atom, 4 O atoms (O4), and 1 Fe atom. On the product (right) side of the equation there are 2H atoms (H2), 3 S atoms (S3), 12 O atoms (O 12), and 2 Fe atoms (Fe2).

For example, the lowest common factor between 2 and 1 is 2 for Fe. Add a 2 in front of the Fe on the left side to balance it. The lowest common factor between 3 and 1 is 3 for S. Add a 3 in front of H2SO4 to balance the left and right sides. At this stage, our equation looks like this: 3 H2SO4 + 2 Fe —> Fe2(SO4)3 + H2

In our example, we added a 3 in front of H2SO4 and now have 6 hydrogens on the left and only 2 on the right side of the equation. We also have 12 oxygen on the left and 12 oxygen on the right, so it is balanced. We can balance hydrogens by adding a 3 in front of H2. Our final balanced equation is 3 H2SO4 + 2 Fe —> Fe2(SO4)3 + 3 H2.

Let’s check our equation, 3 H2SO4 + 2 Fe —> Fe2(SO4)3 + 3 H2, for balance. On the left side of the arrow, there are 6 H, 3 S, 12 O, and 2 Fe. On the right side of the arrow, there are 2 Fe, 3 S, 12 O, and 6 H. The left and the right sides of the equation match, therefore, it is now balanced.

Define the number of atoms of each element in a compound. For example, glucose is C6H12O6, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Identify the atomic mass in grams per mol (g/mol) of each atom. The atomic masses of each element are usually found underneath the element’s symbol on a periodic table, usually as a decimal. The atomic masses of the elements in glucose are: carbon, 12. 0107 g/mol; hydrogen, 1. 007 g/mol; and oxygen, 15. 9994 g/mol. Multiply each element’s atomic mass by the number of atoms present in the compound. Carbon: 12. 0107 x 6 = 72. 0642 g/mol; Hydrogen: 1. 007 x 12 = 12. 084 g/mol; Oxygen: 15. 9994 x 6 = 95. 9964 g/mol. Adding these products together yields the molar mass of the compound. 72. 0642 + 12. 084 + 95. 9964 = 180. 1446 g/mol. 180. 14 grams is the mass of one mole of glucose.

For example: How many moles are in 8. 2 grams of hydrogen chloride (HCl)? The atomic mass of H is 1. 007 and Cl is 35. 453 making the molar mass of the compound 1. 007 + 35. 453 = 36. 46 g/mol. Dividing the number of grams of the substance by the molar mass yields: 8. 2 g / (36. 46 g/mol) = 0. 225 moles of HCl.

For example, what is the molar ratio of KClO3 to O2 in the reaction 2 KClO3 —> 2 KCl + 3 O2. First, check to see the equation is balanced. Never forget this step or your ratios will be wrong. In this case there are equal amounts of each element on both sides of the reaction so it is balanced. The ratio of KClO3 to O2 is 2/3. It doesn’t matter which number goes on top or on bottom as long as you keep the same compounds on the top and bottom throughout the rest of the problem. [11] X Research source

For example, given the reaction N2 + 3 H2 —> 2 NH3 how many moles of NH3 will be produced given 3. 00 grams of N2 reacting with sufficient H2? In this example, sufficient H2 means that there is enough available and you don’t have to take it into account to solve the problem. First, convert grams of N2 to moles. The atomic mass of nitrogen is 14. 0067 g/mol so the molar mass of N2 is 28. 0134 g/mol. Dividing mass by molar mass gives you 3. 00 g/28. 0134 g/mol = 0. 107 mol. Set up the ratios given by the question: NH3: N2 = x/0. 107 mol. Cross multiply this ratio by the molar ratio of NH3 to N2: 2:1. x/0. 107 mol = 2/1 = (2 x 0. 107) = 1x = 0. 214 mol.

The molar mass of NH3 is 17. 028 g/mol. Therefore 0. 214 mol x (17. 028 grams/mol) = 3. 647 grams of NH3.

Generally, a reaction will say that it is given at 1 atm and 273 K or will simply say STP.

For example, convert 3. 2 liters of N2 gas to moles: 3. 2 L/22. 414 L/mol = 0. 143 moles.

The equation can be rearranged to solve for moles: n = RT/PV. The units of the gas constant are designed to cancel out the units of the other variables. For example, determine the number of moles in 2. 4 liters of O2 at 300 K and 1. 5 atm. Plugging in the variables yields: n = (0. 0821 x 300)/(1. 5 x 2) = 24. 63/3. 6 = 6. 842 moles of O2

If the density is not given within the problem, you may have to look it up in a reference text or online.

Identify the volume given. For example, let’s say the problem states that you have 1 liter of H2O. To convert to mL simply multiply by 1000. There are 1000 milliliters in a liter of water.

The density of H2O, for instance, is approximately 1. 0 g/mL.

Define the number of atoms of each element in a compound. For example, glucose is C6H12O6, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Identify the atomic mass in grams per mol (g/mol) of each atom. The atomic masses of the elements in glucose are: carbon, 12. 0107 g/mol; hydrogen, 1. 007 g/mol; and oxygen, 15. 9994 g/mol. Multiply each elements atomic mass by the number of atoms present in the compound. Carbon: 12. 0107 x 6 = 72. 0642 g/mol; Hydrogen: 1. 007 x 12 = 12. 084 g/mol; Oxygen: 15. 9994 x 6 = 95. 9964 g/mol. Adding these products together yields the molar mass of the compound. 72. 0642 + 12. 084 + 95. 9964 = 180. 1446 g/mol. 180. 14 grams is the mass of one mole of glucose.