The answer is 2,511.1023 molecules.
To find the number of molecules, first calculate the amount of O2 in moles using the ideal gas law. Then use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules.
1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP). Therefore, 8.08 L of O2 at STP would contain 8.08/22.4 = 0.36 moles of O2.
The chemical equation for the formation of water from hydrogen and oxygen is: 2H2 + O2 → 2H2O. This equation shows that two molecules of hydrogen gas (H2) react with one molecule of oxygen gas (O2) to form two molecules of water (H2O).
The chemical equation for hydrogen combustion is 2H2 (g) + O2 (g) -> 2H2O (l). This means that two molecules of hydrogen gas react with one molecule of oxygen gas to form two molecules of water.
To calculate the mass of FeO2 produced, we first need to find the number of moles of O2 using the ideal gas law: PV = nRT. At STP, 1 mole of gas occupies 22.4 L, so 50.0 L of O2 is 50.0/22.4 = 2.23 moles. The balanced equation for the reaction will tell you the stoichiometry needed to calculate the mass of FeO2 produced.
To determine the number of moles of MgO produced from 11.2 L of O2, you would first need to balance the chemical equation for the reaction involving MgO and O2. Then, using the ideal gas law and stoichiometry, you can calculate the moles of MgO produced.
22,4 L is the molar volume. 1 mol of oxygen has 32 g and and 6,022140857.10e23 molecules.
1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP). Therefore, 8.08 L of O2 at STP would contain 8.08/22.4 = 0.36 moles of O2.
The chemical equation for the formation of water from hydrogen and oxygen is: 2H2 + O2 → 2H2O. This equation shows that two molecules of hydrogen gas (H2) react with one molecule of oxygen gas (O2) to form two molecules of water (H2O).
The chemical equation for hydrogen combustion is 2H2 (g) + O2 (g) -> 2H2O (l). This means that two molecules of hydrogen gas react with one molecule of oxygen gas to form two molecules of water.
You should set up a balanced equation for the combustion of CH4 first: CH4 + 2 O2 --> CO2 + 2 H2O Now you see that for every 1 mole of CH4 you have 2 moles of O2. Therefore if you have 2.67 L of CH4, you'll have 5.34 L of O2. You can also assume it's at STP just to check, since it all cancels out in the end: 2.67L CH4 (1mol CH4/22.4L)(2mol O2/ 1 mol CH4)(22.4L/1mol O2)= 5.34L O2.
To find the number of hydrogen molecules, first calculate the number of moles in 31.8 L of H2 at STP using the ideal gas law. Then use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules.
55 grams O2 × (1 mol O2 ÷ 32 g O2) × (2 mol H2O ÷ 1 mol O2) × (22.4 L ÷ 1 mol H2O) = 77 L H2O(g)In the liquid form, it works out to just under 62 milliliters.
2 if it has the 2.2 L engine, 3 if it has the 4.3 L engine.
at stp 1 mole of a gas contains 22.4 litres. 9.1/22.4= .40625 moles o2. 1 mole of a gas contains 6.022E23 molecules so .40625 moles x 6.022E23 = 2.4464325E23 molecules, but you have to multiply by two due to it being diatomic, so answer x 2 = 4.892875E23 molecules
To determine the number of moles of MgO produced from 11.2 L of O2, you would first need to balance the chemical equation for the reaction involving MgO and O2. Then, using the ideal gas law and stoichiometry, you can calculate the moles of MgO produced.
One molecule of water consists of two hydrogen atoms and one oxygen atom bonded together.
To calculate the mass of FeO2 produced, we first need to find the number of moles of O2 using the ideal gas law: PV = nRT. At STP, 1 mole of gas occupies 22.4 L, so 50.0 L of O2 is 50.0/22.4 = 2.23 moles. The balanced equation for the reaction will tell you the stoichiometry needed to calculate the mass of FeO2 produced.