Under standard conditions, how many milliliters of hydrogen can be produced by putting 0 23 grams of

Summary. n (h2)＝v/va＝

A certain amount of sodium metal is added to a sufficient amount of loose water, hydrogen gas (standard condition) is generated after full reaction, and the volume of the resulting solution is 200 ml of rubber slag, and the mass of sodium metal consumed and the amount of substance concentration of the obtained solution are calculated.

n (h2)＝v/va＝

2na+2h2o＝2naoh+h2

n(na)＝2（h2）＝

m（na）＝n×m＝

n(naoh)＝2（h2）＝

c（naoh）＝n/v＝

A certain amount of sodium metal is added to a sufficient amount of water, and hydrogen gas (standard condition) is generated after full reaction, and the volume of the obtained solution is 200 ml, and the mass of the sodium metal consumed is calculated as the amount of substance concentration of the obtained solution.

If you can understand it, what else is there you don't understand.

There is no understanding here.

The ratio of the reaction coefficient of Na to H2 is the ratio of the quantity of the substance.

The quantity of the substance, the mass relative to the molecular mass.

2na+2h2o==2naoh+h2↑

xx=2*, so the quality of the inch's spike is the quality of the hydrogen state sleepy early gas.

The amount of substance of Na is n=m m=

2na+2h2o=2naoh+h2↑

n(h2)n(h2)=

The volume of H2 under standard conditions v=n*vm=

o=2naoh+h2

2mol 2mol 2mol

1mol 1mol 1mol 1mol

That is, the hydrogen produced by the Qi Limb Sakura is.

1) A: The volume of hydrogen produced under standard conditions is;

2) the ratio of the number of sodium ions to water molecules in the solution after the reaction is 1; (10-1)=1:9Answer: The ratio of the number of sodium ions to water molecules in the solution after the reaction is 1:9;

Solution: Let the amount of the substance that produces hydrogen be ymol, and the amount of the substance that produces sodium hydroxide is xmol.

2na+2h2o=2naoh+h2↑

x y solves: x=, y=

Hydrogen can be generated, volume =

The amount of sodium hydroxide that produces the substance of the sweet potato = the water that consumes the locust is negligible) Answer...

The amount of HCl in 500ml of 5mol l hydrochloric acid solution 46g of sodium metal 46 23 = 2mol HCL excess, calculated as sodium.

2na＋2hcl=2nacl＋h2↑

2mol xx=

2na＋2hcl→2nacl＋h2

The relative molecular mass of sodium is 23,46 23 2, and the hydrogen produced according to the ratio is 1mol, that is, the space occupied by the gas is 24dm, so hydrogen is generated 24dm

1）1．12l （2）1 mol/l

Question analysis: The equation for the reaction of sodium metal with water is hail oak: 2Na + 2H2O = 2NaOH + H2

n(na)=2 3g 23g mol=0 1mol from the equation n(h2

1 2N(Na)=0 05mol So V(H2=0 05mol 22 4L mol=1 12L. Collapse n(Naoh)=n(Na)=0 1mol C(Naoh)=n(Naoh) V(solution)=0 1mol 0 1L=1mol L. Next to the source.

for, according to the reaction equation:

2na+2h2o=2naoh+h2

So the volume of gas produced is:

The concentration of the substance is: n=