The answer is 1,72 moles.
15 moles of 02 equal 480 g.
The volume of 5.0 moles of 02 at STP is 100 litres.
oxygen
The balanced equation is C3H8 + 5O2 ---> 3CO2 + 4H2O moles C3H8 = 23.7 g x 1 mol/44 g = 0.539 moles moles O2 needed = 5 x 0.539 moles = 2.695 moles O2 (it takes 5 moles O2 per mole C3H8) grams O2 needed = 2.695 moles x 32 g/mole = 86.2 grams O2 needed (3 sig figs)
To determine how many atoms are present in 56 liters of oxygen gas at STP you first need to know that there are two atoms in a single molecule. Then, you would work the scientific formula to determine the number of molecules in the oxygen gas.
15 moles of 02 equal 480 g.
The balanced chemical equation for the reaction between oxygen and hydrogen is2H2 + 02 -> 2H2OThus 2.2 moles of oxygen reacts with 4.4 moles of hydrogen to form 4.4 moles of steam (water in gaseous state).The mass of H2O obtained is thus 4.4 x 18.0 = 79.2g.
1 mole of 02 gas has 12,044 281 714.1023 atoms.
Ten milliliters is a hundredth of a liter. So in a two molar solution, you would have .02 moles in 10 ml.
The volume of 5.0 moles of 02 at STP is 100 litres.
Since we know that one mole of any gas at STP is equal to 22.4 L we can multiply 135L by the following conversion: 1 mole/22.4L. When you set up the problem it looks like this: (135L)x 1 mole/22.4L =6.03 moles of oxygen gas The liters cancel out and you are left with moles as your units. Remember, if you have liters and want moles, divide by 22.4 liters; if you have moles and you want liters you multiply by 22.4 liters.
oxygen
Gas engine has two, diesel has zero.
Molarity = moles of solute/Liters of solutionMoles of solute = Liters of solution * Molarity ( 100 mL = 0.1 Liters )Moles of NaCl = 0.1 Liters * 0.20 M NaCl= 0.02 moles NaCl============
You have to burn C3H8 in O2. You get 3CO2 plus 4H2O. So to burn one mole of C3H8, you need 5 moles of O2. That means you need one fifth of C3H8 as compared to O2. So you need 0.567/5 = 0.1134 moles of C3H8. Hence the answer.
no, but Ozone is
The balanced equation is C3H8 + 5O2 ---> 3CO2 + 4H2O moles C3H8 = 23.7 g x 1 mol/44 g = 0.539 moles moles O2 needed = 5 x 0.539 moles = 2.695 moles O2 (it takes 5 moles O2 per mole C3H8) grams O2 needed = 2.695 moles x 32 g/mole = 86.2 grams O2 needed (3 sig figs)