5.8
The mass of water is 108 g.
0.916 mol
20 moles
0 moles
5048+w4g5=23vm32
2H2 + O2 ---------------> 2H2O for every 2 moles of hydrogen that reacts, 2 moles of water are produced, thus a 1:1 ratio of water produced to hydrogen reacted. So:- 2.5 moles of hydrogen reacted will produce 2.5 moles of water
The answer is: 5 moles oxygen and 4 moles H2O.The reaction is:C3H8 + 5 O2 = 3 CO2 + 4 H2O
1.25xE-3 g/mol O2 reacts completely with 2H2
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
0 moles
5048+w4g5=23vm32
2H2 + O2 ---------------> 2H2O for every 2 moles of hydrogen that reacts, 2 moles of water are produced, thus a 1:1 ratio of water produced to hydrogen reacted. So:- 2.5 moles of hydrogen reacted will produce 2.5 moles of water
The answer is: 5 moles oxygen and 4 moles H2O.The reaction is:C3H8 + 5 O2 = 3 CO2 + 4 H2O
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
1.25xE-3 g/mol O2 reacts completely with 2H2
Well if one mole of water = 2 moles of hydrogen and 1 mole of oxygen, than 2moles of water = 4 moles of hydrogen and 2moles of oxygen.
The answer is o,5 mol.
This answer represents a balanced chemical equation for the combustion of propane (C3H8). When propane reacts with oxygen (O2), it produces carbon dioxide (CO2) and water (H2O).
The equation is 2C4H10 + 13O2 --> 8CO2 + 10H2O This means that for each mole of butane there are 5 moles of water produced. We have 7.01 g of butane = 7.01/58 moles of butane = 0.12 moles. Thus we will get 5 x 0.12 moles of water, = 5 x 0.12 x 18 g of water = 10.88 g.