12 moles Li
Lithium easily react with halogens (group 17).
H3PO4Is phosphoric acid, a strong acid. This would have to react with a strong base, such as NaOH, to produce a salt.3NaOH + H3PO4 --> Na3PO4 + 3H2OThe salt produced is sodium phosphate.
5.0 mol Li * 6.941 g/mol Li = 34.705 = 35 g Lithium
This reaction?? 2H2O -> 2H2 + O2 this would be one to one and you would get; 174.82 moles of water vapor Without an equation I do not really know what you mean.
i think the element will be lithium that's what i think
The nunber of moles of oxygen is 2,5.
Lithium easily react with halogens (group 17).
H3PO4Is phosphoric acid, a strong acid. This would have to react with a strong base, such as NaOH, to produce a salt.3NaOH + H3PO4 --> Na3PO4 + 3H2OThe salt produced is sodium phosphate.
Because 2Li + O2 --> Li2O2 moles Li produces 1 mole Li2O3.04Li + --> 1.52 O2 --> 1.52 Li2O1.52 moles Li2O arw formed.
5.0 mol Li * 6.941 g/mol Li = 34.705 = 35 g Lithium
because it depends on the number of the moles that you will get, so the more moles number that you have for the chlorate the more oxygen that you will get.
lithium, potassium and sodium i think
Iron, lithium, and neon do not actually mix, nor do they chemically react with each other (although iron and lithium react with other elements such as oxygen). Iron is much denser than lithium, so if you poured these two metals into a container in their molten state, the lithium would just float on top of the iron. And Neon is an inert gas.
This reaction?? 2H2O -> 2H2 + O2 this would be one to one and you would get; 174.82 moles of water vapor Without an equation I do not really know what you mean.
i think the element will be lithium that's what i think
2SO2 +O2 -----> 2SO3 so 3.1 moles of SO2 would require 1.55 moles of O2 so since there are 2.7 moles of O2 present and thus in excess, the SO2 is the limiting reagent.
2 Li + Br2 = 2 LiBr is the balanced reaction eq'n. For the second part you need to calculate the moles. moles(Li) = 25 / 7 = 3.57 moles(Br2) = 25 / )80 x 2) = 0.15 BY mathematical equivalence of the reaction eq'n 2:1::2 = 0.3:0.15 :: 0.3 So only 0.3 moles (LI) will be reacted, leaving ( 3.57 - 0.3 = 3.27 moles) unreacted. ( That 22.85 g lithium unreacted) It will give a product mass of 7.2 g (LiBr)