Stoichiometry Calculations PDF

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This document provides a series of calculations and an introduction example of stoichiometry. Calculations in chemistry are used to find the required amounts or products. Examples of calculations include finding the moles of CO2 produced by burning 5 moles of CH4, and how many moles of oxygen are required to form 12 moles of water from the reaction of hydrogen and oxygen. It then further delves into mole-mass calculations for various chemical reactions.

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Calculations from Chemical Equations (Stoichiometry) GENERAL CHEMISTRY 1 MOLE RATIO The combustion of methane, CH4 π‘ͺπ‘―πŸ’ + πŸπ‘ΆπŸ β†’ π‘ͺπ‘ΆπŸ + πŸπ‘―πŸ 𝑢 1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 2 π‘šπ‘œπ‘™π‘’π‘  𝑂2 1 π‘šπ‘œπ‘™π‘’ 𝐢𝑂2 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2...

Calculations from Chemical Equations (Stoichiometry) GENERAL CHEMISTRY 1 MOLE RATIO The combustion of methane, CH4 π‘ͺπ‘―πŸ’ + πŸπ‘ΆπŸ β†’ π‘ͺπ‘ΆπŸ + πŸπ‘―πŸ 𝑢 1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 2 π‘šπ‘œπ‘™π‘’π‘  𝑂2 1 π‘šπ‘œπ‘™π‘’ 𝐢𝑂2 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 2 π‘šπ‘œπ‘™π‘’π‘  𝑂2 2 π‘šπ‘œπ‘™π‘’π‘  𝑂2 2 π‘šπ‘œπ‘™π‘’π‘  𝑂2 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 1 π‘šπ‘œπ‘™π‘’ 𝐢𝑂2 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 MOLE RATIO The combustion of methane, CH4 π‘ͺπ‘―πŸ’ + πŸπ‘ΆπŸ β†’ π‘ͺπ‘ΆπŸ + πŸπ‘―πŸ 𝑢 1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’π‘  𝐢𝑂2 1 π‘šπ‘œπ‘™π‘’π‘  𝐢𝑂2 1 π‘šπ‘œπ‘™π‘’π‘  𝐢𝑂2 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 2 π‘šπ‘œπ‘™π‘’π‘  𝑂2 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 2 π‘šπ‘œπ‘™π‘’π‘  𝑂2 1 π‘šπ‘œπ‘™π‘’ 𝐢𝑂2 MOLE RATIO The combustion of methane, CH4 𝐢𝐻4 + 2𝑂2 β†’ 𝐢𝑂2 + 2𝐻2 𝑂 1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  How many moles of carbon dioxide will be produced by burning 5 moles of methane? π‘₯ π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘‘π‘’π‘ π‘–π‘Ÿπ‘’π‘‘ π‘ π‘’π‘π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘›π‘œ. π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘‘π‘’π‘ π‘–π‘Ÿπ‘’π‘‘ π‘ π‘’π‘π‘ π‘‘π‘Žπ‘›π‘π‘’ (π‘Ÿπ‘’π‘₯ β€² 𝑛) = π‘›π‘œ. π‘œπ‘“ π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘œπ‘‘β„Žπ‘’π‘Ÿ 𝑔𝑖𝑣𝑒𝑛 π‘ π‘’π‘π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘›π‘œ. π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘‘β„Žπ‘’ π‘œπ‘‘β„Žπ‘’π‘Ÿ 𝑔𝑖𝑣𝑒𝑛 π‘ π‘’π‘π‘ π‘‘π‘Žπ‘›π‘π‘’ (π‘Ÿπ‘’π‘₯ β€² 𝑛) MOLE RATIO The combustion of methane, CH4 𝐢𝐻4 + 2𝑂2 β†’ 𝐢𝑂2 + 2𝐻2 𝑂 1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  1 π‘šπ‘œπ‘™π‘’ 2 π‘šπ‘œπ‘™π‘’π‘  How many moles of carbon dioxide will be produced by burning 5 moles of methane? 1 π‘šπ‘œπ‘™π‘’ 𝐢𝑂2 𝑛𝐢𝑂2 1 π‘šπ‘œπ‘™π‘’ 𝐢𝑂2 𝑛𝐢𝑂2 = Γ— 5 π‘šπ‘œπ‘™π‘’π‘  𝐢𝐻4 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 = 5 π‘šπ‘œπ‘™π‘’π‘  𝐢𝐻4 1 π‘šπ‘œπ‘™π‘’ 𝐢𝐻4 𝒏π‘ͺπ‘ΆπŸ = πŸ“ π’Žπ’π’π’†π’” π‘ͺπ‘ΆπŸ MOLE RATIO Determine the number of moles of oxygen reacted with hydrogen to produce 12 moles of water. 2 π‘―πŸ + π‘ΆπŸ β†’ 2 π‘―πŸ 𝑢 𝟐 π’Žπ’π’π’†π’” 𝟏 π’Žπ’π’π’† 𝟐 π’Žπ’π’π’†π’” 𝑛𝑂2 1 π‘šπ‘œπ‘™π‘’ 𝑂2 = 12 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 1 π‘šπ‘œπ‘™π‘’ 𝑂2 𝑛𝑂2 = Γ— 12 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 2 π‘šπ‘œπ‘™π‘’π‘  𝐻2 𝑂 π’π‘ΆπŸ = πŸ” π’Žπ’π’π’†π’” π‘―πŸ 𝑢 How many moles of binary acid must react with a base to produce 15 moles of aluminum bromide in a neutralization reaction. π‘Žπ‘™π‘’π‘šπ‘–π‘›π‘’π‘š π‘π‘Ÿπ‘œπ‘šπ‘–π‘‘π‘’, π΄π‘™π΅π‘Ÿ3 (π‘ π‘Žπ‘™π‘‘) β„Žπ‘¦π‘‘π‘Ÿπ‘œπ‘π‘Ÿπ‘œπ‘šπ‘–π‘ π‘Žπ‘π‘–π‘‘, π»π΅π‘Ÿ(π‘Žπ‘π‘–π‘‘) π‘Žπ‘™π‘’π‘šπ‘–π‘›π‘’π‘š β„Žπ‘¦π‘‘π‘Ÿπ‘œπ‘₯𝑖𝑑𝑒, 𝐴𝑙(𝑂𝐻)3 (π‘π‘Žπ‘ π‘’) πŸ‘π‘―π‘©π’“ + 𝑨𝒍(𝑢𝑯)πŸ‘ β†’ π‘¨π’π‘©π’“πŸ‘ + πŸ‘π‘―πŸ 𝑢 π‘›π»π΅π‘Ÿ 3 π‘šπ‘œπ‘™π‘’π‘  π»π΅π‘Ÿ = 15 π‘šπ‘œπ‘™π‘’π‘  π΄π‘™π΅π‘Ÿ3 1 π‘šπ‘œπ‘™π‘’ π΄π‘™π΅π‘Ÿ3 3 π‘šπ‘œπ‘™π‘’π‘  π»π΅π‘Ÿ π‘›π»π΅π‘Ÿ = Γ— 15 π‘šπ‘œπ‘™π‘’π‘  π΄π‘™π΅π‘Ÿ3 1 π‘šπ‘œπ‘™π‘’ π΄π‘™π΅π‘Ÿ3 𝒏𝑯𝑩𝒓 = πŸ’πŸ“ π’Žπ’π’π’†π’” 𝑯𝑩𝒓 Mole-Mass Calculation Mole to mass Mass to mole Mass to mass Mole to Mass Calculation mole ratio οƒ  mole of desired substance οƒ mass of desired substance Mole-Mass Calculation Mole to mass Mass to mole Mass to mass Mass to Mole Calculation mass to mole οƒ mole ratio οƒ  mole of desired substance Mole-Mass Calculation Mole to mass Mass to mole Mass to mass Mass to Mass Calculation mass to mole οƒ mole ratio οƒ  mole of desired substance οƒ  mass of desired substance Mole-Mass Calculation A complete combustion of 10 moles of ethyl alcohol, C2H5OH, is done producing carbon dioxide and water. Determine the mass of carbon dioxide produced. mole ratio mole of desired substance οƒ mass of desired substance 𝟐π‘ͺ𝟐 π‘―πŸ“ 𝑢𝑯 + πŸ•π‘ΆπŸ β†’ πŸ’π‘ͺπ‘ΆπŸ + πŸ”π‘―πŸ 𝑢 2 π‘šπ‘œπ‘™π‘’π‘  7 π‘šπ‘œπ‘™π‘’π‘  4 π‘šπ‘œπ‘™π‘’π‘  6 π‘šπ‘œπ‘™π‘’π‘  𝑛𝐢𝑂2 4 π‘šπ‘œπ‘™π‘’π‘  𝐢𝑂2 = 10 π‘šπ‘œπ‘™π‘’π‘  𝐢2 𝐻5 𝑂𝐻 2 π‘šπ‘œπ‘™π‘’π‘  𝐢2 𝐻5 𝑂𝐻 𝒏π‘ͺπ‘ΆπŸ = 𝟐𝟎 π’Žπ’π’π’†π’” Mole-Mass Calculation A complete combustion of 10 moles of ethyl alcohol, C2H5OH, is done producing carbon dioxide and water. Determine the mass of carbon dioxide produced. mole ratio mole of desired substance οƒ mass of desired substance 𝟐π‘ͺ𝟐 π‘―πŸ“ 𝑢𝑯 + πŸ•π‘ΆπŸ β†’ πŸ’π‘ͺπ‘ΆπŸ + πŸ”π‘―πŸ 𝑢 𝒏π‘ͺπ‘ΆπŸ = 𝟐𝟎 π’Žπ’π’π’†π’” π‘šπ‘Žπ‘ π‘ πΆπ‘‚2 = 𝑛𝐢𝑂2 Γ— π‘€π‘ŠπΆπ‘‚2 𝑔 π‘€π‘ŠπΆπ‘‚2 = 12 + (16 Γ— 2) π‘šπ‘Žπ‘ π‘ πΆπ‘‚2 = 20 π‘šπ‘œπ‘™π‘’π‘  Γ— 44 𝑔 π‘šπ‘œπ‘™π‘’ π‘€π‘ŠπΆπ‘‚2 = 44 π‘šπ‘œπ‘™π‘’ π’Žπ’‚π’”π’”π‘ͺπ‘ΆπŸ = πŸ–πŸ–πŸŽ π’ˆ Mass to Mole Calculation If 50 grams of sodium is completely synthesized with chlorine to form sodium chloride, how many moles of salt was formed? πŸπ‘΅π’‚ + π‘ͺπ’πŸ β†’ πŸπ‘΅π’‚π‘ͺ𝒍 𝟐 π’Žπ’π’π’†π’” 𝟏 π’Žπ’π’π’† 𝟐 π’Žπ’π’π’†π’” mass to mole mole ratio οƒ  mole of desired substance π‘šπ‘Žπ‘ π‘  π‘π‘Ž π‘›π‘π‘Ž = π‘›π‘π‘Ž = 2.17 π‘šπ‘œπ‘™π‘’π‘  π΄π‘Šπ‘π‘Ž 50 𝑔 π‘›π‘π‘ŽπΆπ‘™ 2 π‘šπ‘œπ‘™π‘’π‘  π‘π‘ŽπΆπ‘™ π‘›π‘π‘Ž = 𝑔 = 23 2.17 π‘šπ‘œπ‘™π‘’π‘  π‘π‘Ž 2 π‘šπ‘œπ‘™π‘’π‘  π‘π‘Ž π‘šπ‘œπ‘™π‘’ 𝒏𝑡𝒂π‘ͺ𝒍 = 𝟐. πŸπŸ• π’Žπ’π’π’†π’”

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