Please help urgent chemistry homework I need the work ?
To Make 44.0 g of carbon dioxide, you must combine 12.0 g of carbon with 32.0 g of oxygen. If a chemist combines 120.0 g of carbon with 160.0 g of oxygen, how many grams of carbon dioxide will be made? If a substance is left over, indicate whether it is carbon or oxygen, and also determine how many grams are left over
回答 (2)
C + O₂ → CO₂
Mass ratio C : O₂ : CO₂ = 12.0 : 32.0 : 44.0
If 120.0 g of carbon completely reacts,
mass of O₂ needed = (120 g) × (32.0/12.0) = 320 g > (Initial mass of O₂)
Hence, O₂ is the limiting reactant
Mass of CO₂ made = (160.0 g) × (44.0/32.0) = 220.0 g
Mass of C left over = (120.0 + 160.0 - 220.0) = 60.0 g
C+O2=CO2..notice the coefficients of the elements in the equation ( 1:1)
120.g or 10 moles of Carbon need 10 mole or 320 g of O2 for a complete reaction
but we have 160.0 g or 5 mole of O2 which reacts with 5 mole or 60 g of C
the C is in excess by 5 mole or 60 g left over.
moles of CO2 product is limited by the limiting reactant .. O2
and the equation shows we get 5 moles of CO2 / mole of O2
5 moles of CO2 = 5 moles * 44 g/ mol = 220 g.
in summary 5 moles of C ( 60g) + 5 moles of O2 ( 160g) give 5 moles of CO2 (220g) leaving 5 moles of C excess
收錄日期: 2021-04-24 08:04:40
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