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How many moles of O2 are in 8.25 g of O2?


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Answered 2016-08-18 20:22:27

The answer is 4,09 moles.

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150.0 g O2 x 1 mole O2/32 g O2 = 4.688 moles O2


2H2 (g) + O2 (g) ---> 2H20 (l)8 moles H2 + 4 moles O2 --->8 moles H2O so 4 moles of O2


0.50 moles O2 in 16 g (2 significants)


C3H8 + 5O2 ==> 3CO2 + 4H2O balanced equationmole ratio O2:C2H8 = 5:1 1.5 moles C3H8 x 5 moles O2/mole C3H8 = 7.5 moles O2 needed


The balanced equation is C3H8 + 5O2 ---> 3CO2 + 4H2O moles C3H8 = 23.7 g x 1 mol/44 g = 0.539 moles moles O2 needed = 5 x 0.539 moles = 2.695 moles O2 (it takes 5 moles O2 per mole C3H8) grams O2 needed = 2.695 moles x 32 g/mole = 86.2 grams O2 needed (3 sig figs)


16 g O2 correspond to 0,5 mol.


2SO2 + O2 ==> 2SO3moles of SO2 present = 98.6 g x 1 mole/64 g = 1.54 molesmoles O2 needed = 1/2 (1.54 moles) = 0.770 moles O2 needed(b/c mole ratio of O2:SO2 is 1:2)


4 NH3 + 5 O2 ---> 4 NO + 6 H2Omoles NH3 used = 36.3 g x 1 mole/17 g = 2.14 molesmoles O2 needed = 2.14 moles (note a 1mole to 1mole ratio of O2 to NH3 in balanced equation)grams O2 needed = 2.14 moles x 32g/mol = 68.48 grams needed


0,800 moles of oxygen (O2) is equivalent to 25,6 g.


1CO2 ==> 1O26.5 g O2 x 1 mol/32 g = 0.203 moles O2grams of CO2 = 0.203 moles x 44 g/mole = 8.9 grams


N2O5(g) → 4NO2(g) + O2(g)


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)


The reaction of carbon dioxide and potassium oxide is 4KO2 + 2CO2 = 2K2CO3 + 3O2. 156 grams of CO2 is 3.54 moles, which will produce 5.31 moles of O2.


Moles Mg = 3.00 g / 24.312 g/mol =0.123 Moles O2 = 2.20 / 32 g/mol = 0.0688 2 Mg + O2 >> 2 MgO the ratio between Mg and O2 is 2 : 1 0.123 / 2 = 0.0615 moles O2 needed we have 0.0688 moles of O2 so O2 is in excess and Mg is the limiting reactant we get 0.123 moles of MgO => 0.123 mol x 40.31 g/mol =4.96 g



2 N2 (g) + O2 (g) ⇌ 2 N2O (g)16 g O2 x 1 mole O2/32 g = 0.5 moles O2 present 0.5 moles O2 x 2 moles N2/mole O2 = 1.0 mole N2 required Mass of N2 required = 1.0 mole x 28 g/mole = 28 grams N2 required.


Balanced equation. 2H2 + O2 -> 2H2O 355 grams O2/32 grams = 11.1 moles O2 check for limiting reactant 11.1 moles O2 (2 mole H2/1 mole O2) = 22.2 mole H2 and H2 has no where near that many moles, so limits and drives reaction so, as they are one to one...... 22.2 moles of H2O are produced


1 mole O2 weighs 32 gso 64 g O2 is 2 mol O2 gaswhich has 2 X 6.022 X 1023 = 1.2066 X 1024 molecules of O2


The number of moles in 24 g of oxygen gas (O2) is 0,75.


Ethane is C2H6 and combustion of ethane is 2C2H6 + 7O2 ==> 4CO2 + 6H2OThus for combustion of 2 moles of ethane, one requires 7 moles O2The mass of O2 required thus is 7 moles O2 x 32 g/mole O2 = 224 g of O2 needed


24 g of oxygen gas (O2) is equal to o.75 moles.




The mass of carbon dioxide is 220,5 g.


You have :N2 + O2 -----> 2NOmoles NO = ( 2 mols N2 reacted ) ( 2 mol NO / mol N2 reacted )moles NO = 4 moles NOmass NO = ( 4 mol NO ) ( 30 g NO / mol NO ) = 120 g NO



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