85 moles Li x 6.02x10^23 atoms/mole Li = 5.17x10^25 atoms of Li
To determine the number of moles of water needed to react with 1.7 moles of Li2O, we can use the balanced chemical equation for the reaction: [ \text{Li}_2\text{O} + 2\text{H}_2\text{O} \rightarrow 2\text{LiOH}. ] From the equation, 1 mole of Li2O reacts with 2 moles of water. Therefore, 1.7 moles of Li2O would require 1.7 x 2 = 3.4 moles of water.
1 mole Li = 6.94g Li = 6.022 x 1023 atoms Li 27.0g Li x 6.022 x 1023 atoms Li/6.94g Li = 2.34 x 1024 atoms Li
20,32 g of lithium nitride can be obtained.
3.977 mol
1 mole of Li₂O contains 2 moles of lithium (Li) atoms and 1 mole of oxygen (O) atoms. Therefore, in 1 mole of Li₂O, there are a total of 3 moles of atoms.
85 moles Li x 6.02x10^23 atoms/mole Li = 5.17x10^25 atoms of Li
From the periodic table, lithium has an atomic weight of 6.941. The molar mass of an element is the atomic weight in grams. Therefore, 1 mole Li = 6.941g Li Therefore, moles Li = 15g Li X 1 mole Li/6.941g Li = 2.2 moles Li
4.7 mole Li (6.022 X 10^23/1 mole Li) = 2.8 X 10^24 atoms Li
The balanced chemical equation for the reaction is 3Li + H3PO4 -> Li3PO4 + 3H2. The mole ratio is 1:1 between Li and H3PO4. Therefore, 4 moles of H3PO4 will react with 4 moles of Li.
3,00 moles of Li have 18,066422571.10e23 atoms.
(7.6g LiBr)/(86.84g/mol) x (1molLi/1molBr) = .0875 mol Li or with sig figs, .088
This reaction? 6Li + N2 --> 2Li3N 0.450 moles Li (2 moles Li3N/6 moles Li) = 0.150 moles lithium nitride produced ===========================
15g Li * (1mol Li / 6.941g Li) = 2.16 mol Li
For this problem, the atomic mass is not required. Take the mass in moles and multiply it by Avogadro's constant, 6.02 × 1023. Divide by one mole for the units to cancel.2.5 moles H2 × (6.02 × 1023 atoms) = 1.51 × 1024 atoms
I assume you mean 97.9 grams lithium. 97.9 grams lithium (1 mole Li/6.941 grams)(6.022 X 10^23/1 mole Li) = 8.49 X 10^24 atoms of lithium ---------------------------------------
No moles of BaCO3 are found in any amount of Li.