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.
From the balanced chemical equation, 1 mole of C5H12 reacts with 8 moles of O2 to produce 6 moles of H2O. Therefore, when 1 mole of O2 reacts, it will produce 6 moles of 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.
The molecular mass of O2 (oxygen gas) is 32 g/mol. This is calculated by adding the atomic masses of the two oxygen atoms in the molecule (16 g/mol each).
Type your answer here... 2 × (6.02 × 1023)
When gasoline is heated in the presence of hydrogen gas and a catalyst, the gasoline crack. The cracking gasoline decomposes to 1 mol of methane, 2 mol of ethane, and 1 mole of propane for jet fuel. It is a process known as hydrocracking.
R = 8.314472(15) J K−1 mol−1
1 mol of any gas has a volume of 22.4 L at STP
Using the following thermochemical data, what is the change in enthalpy for the following reaction: 3H2(g) + 2C(s) + ½O2(g) → C2H5OH(l) C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l), ΔH = –1367 kJ/mol C(s)+O2(g)→CO2(g), ΔH = –393.5 kJ H2(g)+½O2(g)→H2O(l), ΔH = –285.8 kJ a. -277.6 kJ/mol b. -194.7 kJ/mol c. 194.7 kJ/mol d. 486 kJ/mol
16 a.m.u or 16 g mol-1
1) single replacement reaction: Cl2 (g) + 2KI (s) --> 2KCl (s) + I2 (g) 2) correctly balanced equation: 4Al (s) + 3O2 (g) --> 2Al2O3 (s) 3) liters of hydrogen gas are required to produce 2 moles of NH3 2mol NH3 = 44.8 litres at STP 3litres N2 produce 2 litres NH3 vol H2 = 44.8*3/2 = 67.2 litres N2 required 4)grams of oxygen are required to produce 1 mole of H2O? 3mol O2 produce 6mol H2O 0.5mol O2 will produce 1 mol H2O Molar mass O2 = 32g/mol 0.5mol = 16g O2
22,4 L is the molar volume. 1 mol of oxygen has 32 g and and 6,022140857.10e23 molecules.
6.65X10^5 kj/mol
Assuming that this ammonia gas is at STP, you can use Avogadro's number to gind the number of moles of gas:(387 x 1021 molecules) x (1 mol / 6.02x1023particles) x (17.03 g / 1 mol) =110 g NH3
To determine the number of molecules of a substance, we can use Avogadro's number, which states that in every mole of a substance, there are 6.022 x 10^22 particles, or in this case molecules. This can be rewritten as the ratio: 1 mol / 6.022 x 10^22 molecules.This problem is deceptively tricky: you might think that given 3 moles oxygen, we can just write it as 3 mol O. However, it is important to note that oxygen naturally exists as O2. So, we have to convert what's given into moles of O3.00 mol O2 x 2 mol O = 6.00 mol O1 mol O2In this problem, we want to cancel out the mol unit in the numerator. We can flip the ratio above to make it 6.022 x 10^23 molecules/1 mol6.00 mol O x 6.022x10^22 molecules O = 3.61x10^22 molecules O1 mol O
4.3 g * 1 mol/18.02 g * 6.03 kJ/mol