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The reaction between Isopropyl alcohol and oxygen is 2 C3H8O + 9 O2 equals 6 CO2 + 8 H2O. So for every mole of isopropyl alcohol, 4.5 moles of oxygen are consumed. 6.5 grams of C3H8O is .108 moles and 12.3 grams of O2 is .384 moles. This means that O2 is the limiting reactant as it needs .486 moles of O2 to finish.
The total mass in a chemical reaction remain unchanged: 50 g.
5a+10b = 10ab
2Mg + O2 -> 2MgO1 mole of Oxygen yields two moles of MgOmoles of oxygen = 0.643 molmoles of MgO formed = 1.286 molgrams of MgO formed = 40.3 X 1.286 = 51.8258 g
how many grams of oxygen are consumed when 19.4g of carbon dioxide is formed during the combustion of C7H16
42.1g/ 59g/mol (grams/molar mass of C3H8O)
The reaction between Isopropyl alcohol and oxygen is 2 C3H8O + 9 O2 equals 6 CO2 + 8 H2O. So for every mole of isopropyl alcohol, 4.5 moles of oxygen are consumed. 6.5 grams of C3H8O is .108 moles and 12.3 grams of O2 is .384 moles. This means that O2 is the limiting reactant as it needs .486 moles of O2 to finish.
25
99.8gS
The answer is: 4,2 g propene and 1,8 g water; it is a dehydration reaction.
234 grams
340 grams
If the reaction is:6 Na + 2 O2 = 2 Na2O + Na2O2This mass is 3,83 g sodium.
133
62 grams a+
62 grams a+
You can't answer this question unless you know the the NO2 was formed FROM. You need to write the balanced reaction for the reaction and then use stoichiometry to solve for the amount of oxygen produce.See the Related Questions to the left for how to write a balanced reaction and how to use stoichiometry to solve this type of problem.