38 grams oxygen gas (1 mole O2/32 grams)(6.022 X 1023/1 mole O2)
= 7.2 X 1023 molecules of oxygen gas
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3.120 grams oxygen gas (1 mole O2/32 grams)(6.022 X 1023/1 mole O2) = 5.871 X 1022 atoms of oxygen gas ------------------------------------------------
When you resubmit your question please indicate which gas is of interest to you.
The answer is 6,992 g oxygen.
6H2 + 6O2 ---> 6H2O + 3O2 This reaction leave excess unreacted oxygen behind.
The number of molecules in 15 g ethane is approx. 3,011.10e23.
9.03 × 1023
1.659 [grams] / 47.998 [grams / mol] * 6.02214179(30)×1023 [molecules / mole] * 3 [atoms / molecule]. A bunch.
For what purpose?
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.
3.120 grams oxygen gas (1 mole O2/32 grams)(6.022 X 1023/1 mole O2) = 5.871 X 1022 atoms of oxygen gas ------------------------------------------------
It is not measured in grams but PPM because it is a gas.
When you resubmit your question please indicate which gas is of interest to you.
That depends on the fuel. Natural gas, methane, needs 2 molecules of oxygen, but gasoline, octane, needs 17 molecules of oxygen.
The answer is 6,992 g oxygen.
6,5 moles oxygen equals 208 g.
6H2 + 6O2 ---> 6H2O + 3O2 This reaction leave excess unreacted oxygen behind.
Assuming you mean oxygen gas, the number of molecules can be found by first finding the number of moles = mass of oxygen (4g) / Molecular mass of oxygen gas (32 g mol-1) This tells us there is 0.125 mol of oxygen gas present. The number of molecules present is given by the number of moles x the avogadro constant (6.022x10^23) So the number of oxygen gas molecules present is equal to 0.125 x 6.022x10^23 = 7.5275x10^22 molecules