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This depends on the proportion of these gases in the reactor.
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.
Methane reacts with oxygen in the following way. CH4 + 3 O2 --> CO2 + 4 H2O. If 5 moles of oxygen react with 2.8 moles of methane, only 1.67 moles of methane would be consumed because of the molar ratio 1:3. This would produce 1.67 moles of carbon dioxide and 6.67 moles of water.
TiCl4 is limiting reagent, O2 is in excess
The gas that is a reactant in aerobic cellular respiration is oxygen. C6H12O6 + O2 --> CO2 + H2O
The limiting reactant is oxygen.
When 85.0 g of CH4 are mixed with 160. g of O2 the limiting reactant is __________. CH4 + 2O2 → CO2 + 2H2O
This depends on the proportion of these gases in the reactor.
Cus
There is no limiting reactant in that equation, it's balanced. Four hydrogens on the left, 4 on the right, 2 oxygens on the left, 2 oxygens on the right. If it was 3H2 then it would be oxygen.
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.
I think that should be 2N2O4+2H2O+O2 --> 4HNO3 But i'm not sure
There is no limiting reactant in that equation, it's balanced. Four hydrogens on the left, 4 on the right, 2 oxygens on the left, 2 oxygens on the right. If it was 3H2 then it would be oxygen.
Moles Mg = 3.00 g / 24.312 g/mol =0.123 Moles O2 = 2.20 / 32 g/mol = 0.0688 2 Mg + O2 >> 2 MgO the ratio between Mg and O2 is 2 : 1 0.123 / 2 = 0.0615 moles O2 needed we have 0.0688 moles of O2 so O2 is in excess and Mg is the limiting reactant we get 0.123 moles of MgO => 0.123 mol x 40.31 g/mol =4.96 g
Methane reacts with oxygen in the following way. CH4 + 3 O2 --> CO2 + 4 H2O. If 5 moles of oxygen react with 2.8 moles of methane, only 1.67 moles of methane would be consumed because of the molar ratio 1:3. This would produce 1.67 moles of carbon dioxide and 6.67 moles of water.
In order to have a balanced equation, adding the reactant H2 And O2 (H2+O2) have to come out equal on the product side and the reactant side. This would look like H2+O2 = H2O2
TiCl4 is limiting reagent, O2 is in excess