Hydrated Iron (III) oxide:
2Fe + 1.5O2 + xH2O -> Fe2O3·xH2O
where x is the number of H2O molecules present.
Cu (copper) reacts with CO2 (carbon dioxide), H2O (water), and O2 (oxygen) to form a product. The specific product formed depends on the reaction conditions and stoichiometry. Without further information, it is not possible to determine the exact reaction and product formed.
The product is water:2 H2 + O2 = 2 H2O
First off, you decide the product (becomes easier after a while of doing chemistry). The product is H2O (water). O2+H2=>H2O, but this is not stochiometrically balanced, so you have to change the amount of H2O's on the right side of equation and then to balance the amount of hydrogens. If you add a 2 in front of both, you get O2 + 2H2 => 2H2O
C5H12 + 8 O2 --> 5 CO2 + 6 H2O (1 mol O2)(6 mol H2O/8 mol O2) = 0.75 mol H2O
This equation is:C2H5OH + 3 O2 = 2 CO2 + 3 H2O
Cu (copper) reacts with CO2 (carbon dioxide), H2O (water), and O2 (oxygen) to form a product. The specific product formed depends on the reaction conditions and stoichiometry. Without further information, it is not possible to determine the exact reaction and product formed.
The product is water:2 H2 + O2 = 2 H2O
The product of the reaction between hydrogen peroxide (H2O2) and manganese dioxide (MnO2) is oxygen gas (O2) and water (H2O).
When hydrogen gas (H2) reacts with oxygen gas (O2), they combine to form water (H2O) as a product. This reaction releases energy in the form of heat and light.
The coefficient of O2 is 5.The chemical equation is:C5H12 + 8 O2 = 5 CO2 + 6 H2O
2NaH2 + O2 yields 2Na + 2H2O
First off, you decide the product (becomes easier after a while of doing chemistry). The product is H2O (water). O2+H2=>H2O, but this is not stochiometrically balanced, so you have to change the amount of H2O's on the right side of equation and then to balance the amount of hydrogens. If you add a 2 in front of both, you get O2 + 2H2 => 2H2O
C2H6 + O2 = CO2 + H2O There are two carbons in the reactant, so there are two carbon in 'CO2', by writing '2CO2' . Hence C2H6 + O2 = 2CO2 + H2O Similarly, there are six hydrogen in the reactant, so there are six hundreds in H2O , by writing ' 3H2O . Hence C2H6 + O2 = 2CO2 + 3H2O To balance the oxygens, we now think ' backwards'. We note in the products there are 2 x 2 = 4 oxygen in CO2. and 3 x 1 = 3 in H2O. So we need 4 + 3 = 7 oxygens in the reactants. We already have '2' oxygen in the reactants as 'O2' . To make this up to '7' , we write ' 3 1/2' as the molar ratio/. ( 2 x 3 1/2 = 7) Hence C2H6 + ( 3 1/2)O2 = 2CO2 + 3H2O Chemists do NOT like fractions in the molar ratios. So to remove the '3 1/2' we multiply the whole reaction eq'n by '2'. Hence 2C2H6 + 7O2 = 4CO2 + 6H2O is the balanced reaction eq'n. NB Note the number of each type of atom are equal on both sides. 4 carbons 12 hydrogens 14 oxygens. NNB When doing these combustion eq'ns for alkanes, alkenes, alkynes, etc., Think to forward balance the carbons and hydrogens, then 'Back' think of the number of oxygens.
The unbalanced combustion reaction of C4H10(g) with O2(g) produces CO2(g) and H2O(g) as products. The balanced reaction is: C4H10(g) + O2(g) → CO2(g) + H2O(g)
This simple equation is: 2H2 + O2 = 2H2O
C5H12 + 8 O2 --> 5 CO2 + 6 H2O (1 mol O2)(6 mol H2O/8 mol O2) = 0.75 mol H2O
The reaction is:4KO2 + 2H2O ---> 4KOH + 3O2