One mole of O₂ gas occupies approximately 22.4 liters at standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atmosphere of pressure. The molar mass of O₂ is about 32 grams, meaning that 1 mole of O₂ weighs 32 grams. Thus, under STP conditions, 1 mole of O₂ gas will have both a specific volume and mass.
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.
When comparing 1 mol of oxygen gas (O₂) with 1 mol of carbon monoxide gas (CO), the number of molecules must be the same, as both contain 1 mol. Additionally, both gases will have the same volume under standard temperature and pressure conditions, which is approximately 22.4 liters. However, their molar masses and chemical properties differ significantly.
C5H12 + 8 O2 --> 5 CO2 + 6 H2O (1 mol O2)(6 mol H2O/8 mol O2) = 0.75 mol H2O
The balanced equation is: 2C + O2 -> 2CO2. First, determine the moles of C and O2: 4g C / 12 g/mol = 0.33 mol C and 10.67g O2 / 32 g/mol = 0.33 mol O2. From the balanced equation, 2 moles of C produces 2 moles of CO2, so 0.33 mol C will produce 0.33 mol CO2. Since CO2 has a molar mass of 44 g/mol, the total grams of CO2 produced will be: 0.33 mol CO2 x 44 g/mol = 14.52 grams of CO2.
To calculate the number of oxygen atoms in 16.0 pounds of oxygen, first convert 16.0 pounds to grams (1 pound ≈ 453.592 grams). Then, calculate the number of moles of oxygen using the molar mass of oxygen (16.00 g/mol). Finally, use Avogadro's number (6.022 x 10^23 atoms/mol) to find the number of oxygen atoms.
15.99 x 2 g
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.
Using the molar mass of oxygen (O2) which is 32 g/mol, you can calculate that you have 3 moles of oxygen gas. Since the ratio of O3 to O2 is 1:2, you would form 1.5 moles of ozone (O3) from 96.0g of O2.
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
By using the balanced chemical equation for the decomposition of mercury(II) oxide (HgO): 2 HgO -> 2 Hg + O2, we see that 1 mol of HgO produces 1 mol of O2. Therefore, 0.437 mol of HgO will produce 0.437 mol of O2. To convert mol to grams, we use the molar mass of oxygen: 32.00 g/mol. so, 0.437 mol of O2 is equivalent to 0.437 mol * 32.00 g/mol = 13.92 grams of O2.
Do you mean this reaction? 2Mg + O2 -> 2MgO 4.11 moles Mg (1 mole O2/2 mole Mg) = 2.06 moles oxygen gas consumed --------------------------------------------------
4H2O2 --> 4H2O + 2O2 That would be 2: one mol of O2 for every 2 mols of H2O2.
Using the balanced chemical equation: 2Mg + O2 -> 2MgO, we can see that 1 mol of O2 reacts with 2 moles of Mg to produce 2 moles of MgO. Calculate the moles of O2: 40g O2 / 32g/mol = 1.25 mol O2 From the equation, 1.25 mol O2 will produce 2.5 mol of MgO. Calculate the grams of MgO produced: 2.5 mol MgO * 40.3 g/mol = 100.75 grams MgO.
1 mol of octane (C8H18) reacts with 25 mol of oxygen (O2) to produce 8 mol of CO2 and 9 mol of H2O. Therefore, 0.74 mol of oxygen can react with 0.0744 mol of octane.
Type your answer here... 2 × (6.02 × 1023)
2Mg + O2 ==> 2MgO Balanced Equation4.03 g Mg x 1 mole Mg/24.3 g = 0.166 moles Mg present in 4.03 g Mg.moles O2 required = 0.166 moles Mg x 1 mole O2/2 moles Mg = 0.083 moles O2 needed.At STP 1 mole occupies 22.4 L, thusVolume of O2 required = 0.083 moles x 22.4 L/mole = 1.56 L x1000 ml/L = 1859 mlsSince 4.03 g has only 3 significant figures, the correct answer should be 1860 milliliters.
C5H12 + 8 O2 --> 5 CO2 + 6 H2O (1 mol O2)(6 mol H2O/8 mol O2) = 0.75 mol H2O