CHEM 4211-How many grams of carbon dioxide can form when a mixture of 4.41

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June 8th, 2024

How many grams of carbon dioxide can form when a mixture of 4.41 g propane (C3H8) and 6.40 g of oxygen (O2) is ignited, assuming complete combustion to form carbon dioxide and water. The balanced equation is:C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)Question 1 options:1.99 g10.8 g13.2 g5.28 gSaveQuestion 2 (1 point)This is a continuation of a previous problem. A mixture of 4.41 g propane (C3H8) and 6.40 g of oxygen (O2) is ignited. What reactant is in excess and by how much assuming complete combustion to form carbon dioxide and water. The balanced equation is:C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)Question 2 options:O2 is in excess by 1.99 gC3H8 is in excess by 1.99 gO2 is in excess by 1.76 gC3H8 is in excess by 2.65 gSaveQuestion 3 (1 point)The combustion of butane produced 0.0243 kg of carbon dioxide. What mass of butane is required if the yield is 89 %.Question 3 options:9.01 g36.1 g144.2 gSaveQuestion 4 (1 point)Hydrogen and oxygen react to form water. Consider the mixture of hydrogen and oxygen shown below. The filled circles represent oxygen and the open circles represent hydrogen. Select the correct statement. Question 4 options:H2 is in excess by 3 moleculesH2 is in excess by 1 moleculeH2 is in excess by 2 moleculesH2 is in excess by 5 molecules

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