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Mass to Mass Stoichiometry. Honors Chemistry Unit 5. Ex. How many grams of copper (I) sulfide could be produced from 7.90 g copper (I) chloride reacting with an excess of H2S gas? Steps to Solve : 1. Write a balanced equation. Determine what you start with and what you want to know.
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Mass to Mass Stoichiometry Honors Chemistry Unit 5
Ex. How many grams of copper (I) sulfide could be produced from 7.90 g copper (I) chloride reacting with an excess of H2S gas? • Steps to Solve: 1. Write a balanced equation. Determine what you start with and what you want to know. H2S + 2CuCl Cu2S +2HCl
H2S + 2CuCl Cu2S +2HCl 2. Convert grams to moles. 7.90 g CuCl 1mole CuCl = 0.0798 mol CuCl 99.0 g CuCl 3. Covert moles of what you start with to moles of what you want to know using a mole ratio. You get this ratio from the balanced chemical reaction! 0.0798 mol CuCl 1 mol Cu2S = 0.0399 mol Cu2S 2 mol CuCl 4. Convert moles of what you want to know to grams. 0.0399 mol Cu2S 159.1 g Cu2S = 6.35 g Cu2S 1 mol Cu2S
This can also be done all together as follows: 7.90 g CuCl 1mole CuCl 1 mole Cu2S 159.1g Cu2S 99.0 g CuCl 2 moles CuCl 1mole Cu2S
2NaOH + H2SO4 2H2O + Na2SO4 How many grams of sodium sulfate will form if you start with 200. grams of sodium hydroxide and you have an excess of sulfuric acid? 1 mole NaOH 1 mole Na2SO4 200 g NaOH 142.05 g Na2SO4 40.0 g NaOH 2 moles NaOH 1 mole Na2SO4 = 355.1 grams Na2SO4
Pb(SO4)2 + 4LiNO3 Pb(NO3)4 + 2Li2SO4 How many grams of lithium nitrate will be needed to make 250. grams of lithium sulfate, assuming you have enough lead (IV) sulfate to complete the reaction? 1 mole Li2SO4 4 moles LiNO3 68.95 g LiNO3 250.g Li2SO4 109.95 g Li2SO4 2 moles Li2SO4 1 mole LiNO3 = 313.6 grams LiNO3
Other Examples 2KClO3 2KCl + 3O2 Given 0.0032 moles of KClO3, how many grams of KCl will form? 0.0032 moles KClO3 2 moles KCl 74.55 g KCl 2 moles KClO3 1 mole KCl 0.239 grams KCl will form
2PbS+ 3O2 2PbO + 2SO2 Given 6.64 x 10-3gramsof oxygen, how many moles of SO2 will form? 6.64 x 10-3 g O2 1 mole O2 2 moles SO2 32.0 g O2 3 moles O2 = 1.38 x 10-4 moles SO2