Sodium and water react according to the equation 2na(s) + 2 h2o (liquid) -> 2 naoh (aq) + h2 (g). if 10.0 g id sodium is placed in 50.0 g of water, calculate the mass of hydrogen that will be produced.

2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)

1) 2 reactants are given, so we have to find which of them is an excess reactant and which of them will react completely.

Convert gram of Na and H2O into moles.

M(Na) =23.0 g/mol

10.0 g Na*1 mol Na/23.0 g Na≈0.4349 mol Na

M(H2O) = 2*1.0 +16.0 = 18.0 g/mol

50.0 g H2O*1 mol H2O/18.0 g H2O =2.778 mol H2O

2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)

from equation 2 mol 2 mol

given 0.4349 mol 2.778 mol

We can see that H2O is an excess reactant, so we are going to find amount of H2 using Na.

2)

2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)

from equation 2 mol 1 mol

given 0.4349 mol x mol

x=(0.4349*1)/2 =0.21745 mol H2

3) mass of H2

0.21745 mol H2 * 2.0 g H2/1 mol ≈ 0.435 g H2

Answer : 0.435 g H2.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.