2.5 moles H2O (2 moles H/1 mole H2O)
= 5 moles of hydrogen
For this you need the atomic (molecular) mass of H2O. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. H2O= 18.0 grams2.5 moles H2O × (18.0 grams) = 45.0 grams H2O
Balanced equation first. 2H2 + O2 -> 2H2O 250 moles O2 (2 mole H2O/1 mole O2) = 500 mole H2O produced now, since I am forgetful, I will use density formula at 25 C Density = grams/milliliters 500 moles H2O (18.016 grams/1 mole H2O) = 9008 grams H2O 0.9982 g/ml = 9008 grams/milliliters 9024.24 milliliters H2O this is...... 9.02 liters of water produced in this reaction.
2H2 + O2 ===> 2H2O50 grams H2 x 1 mole H2/2 g = 25 moles H2300 grams O2 x 1 mole O2/32 g = 9.375 moles O2Limiting reactant is O2, so maximum moles of H2O formed = 9.375 O2 x 2H2O/1 O2 = 18.75 molesgrams H2O = 18.75 moles H2O x 18 g/mole = 337.5 g H2O = 340 g (to 2 significant figures)
301 grams H20 ( 1mol H2O/18.016g )(2mol H/1mol H2O )(6.022 X 10^23/1mol H ) = 2.01 X 10^25 atoms of hydrogen
360 g H2O x 1 mol/18g H2O x 6.02x10^23 H2O molecules/mol x 2 H atoms/molecule = 2.4x10^25 hydrogen atoms.
25x(2/18)=2.78g of hydrogen 25x(16/18)=22.22g of oxygen
For this you need the atomic (molecular) mass of H2O. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. H2O= 18.0 grams2.5 moles H2O × (18.0 grams) = 45.0 grams H2O
Balanced equation first. 2H2 + O2 -> 2H2O 250 moles O2 (2 mole H2O/1 mole O2) = 500 mole H2O produced now, since I am forgetful, I will use density formula at 25 C Density = grams/milliliters 500 moles H2O (18.016 grams/1 mole H2O) = 9008 grams H2O 0.9982 g/ml = 9008 grams/milliliters 9024.24 milliliters H2O this is...... 9.02 liters of water produced in this reaction.
2H2 + O2 ===> 2H2O50 grams H2 x 1 mole H2/2 g = 25 moles H2300 grams O2 x 1 mole O2/32 g = 9.375 moles O2Limiting reactant is O2, so maximum moles of H2O formed = 9.375 O2 x 2H2O/1 O2 = 18.75 molesgrams H2O = 18.75 moles H2O x 18 g/mole = 337.5 g H2O = 340 g (to 2 significant figures)
Balanced equation: 2C8H18 + 25O2 ==> 16CO2 + 18H2Omoles of octane used: 325 g x 1 mole/114g = 2.85 moles octanemoles H2O produced: 18 moles H2O/2 moles C8H18 x 2.85 moles C8H18 = 25.65 moles H2O
Balanced equation. 2C8H18 + 25O2 -> 16CO2 + 18H2O 5.5 moles C8H18 (25 moles O2/2 moles C8H18) = 68.75 moles O2 needed
301 grams H20 ( 1mol H2O/18.016g )(2mol H/1mol H2O )(6.022 X 10^23/1mol H ) = 2.01 X 10^25 atoms of hydrogen
Molarity = moles of solute/Liters of solution Molarity = 25 moles sucrose/50 liters H2O = 0.5 M sucrose
360 g H2O x 1 mol/18g H2O x 6.02x10^23 H2O molecules/mol x 2 H atoms/molecule = 2.4x10^25 hydrogen atoms.
There are four molecules of water in 4H2O. One molecule of water is written as H2O. The subscript 2 tells you that there are two H(hydrogen) atoms. Since there is no subscript after the O there is one O(oxygen) atom. Now back to 4H2O. The coefficient 4 tells you there are four molecules of H2O. Therefore there are eight H(hydrogen) atoms and four O(oxygen) atoms.
How many hydrogen atoms are in 35.0 grams of hydrogen gas?
.071 moles