Stoichiometry question help

[abelkoontz]

First of all, when you have 2Na and you try to find the moles, do you just disregard the 2? Because the mols is 0.05 right? 

So 2NaOH also has 0.05mol. And i'm stuck lol

[SC2Player]

The coefficients specify the ratios in which the reactants react and the products are produced.  In this case, Na reacts with water in a 2:2 ratio to produce NaOH and H2 in a 2:1 ratio, and so the overall reaction ratio is in 2:2:2:1.  Hence, 0.05 mol of Na reacts with 0.05 mols of water, producing 0.05 mols of NaOH and 0.025 mols of H2.  From thereon, you can calculate the concentration of the NaOH solution.  

[abelkoontz]

13 minutes ago, SC2Player said:

The coefficients specify the ratios in which the reactants react and the products are produced.  In this case, Na reacts with water in a 2:2 ratio to produce NaOH and H2 in a 2:1 ratio, and so the overall reaction ratio is in 2:2:2:1.  Hence, 0.05 mol of Na reacts with 0.05 mols of water, producing 0.05 mols of NaOH and 0.025 mols of H2.  From thereon, you can calculate the concentration of the NaOH solution.  

 

How do you calculate the conc? Do you just pretend volume = mass 

[SC2Player]

Concentration is amount of something per unit volume (hence units mol dm^-3).  Take amount of NaOH you have in mols and divide this by volume (250 cm^3), then simplify into mol dm^-3.

[kw0573]

1 hour ago, abelkoontz said:

First of all, when you have 2Na and you try to find the moles, do you just disregard the 2? Because the mols is 0.05 right? 

So 2NaOH also has 0.05mol. And i'm stuck lol

Yes, the 2 in front of Na (the coefficient) does not matter. The important detail is the ratio moles of NaOH produced / moles of Na consumed, which is unity (1). That means with 0.05 mol of Na consumed or 0.05 mol of NaOH produced, in 0.250 dm³ water, the concentration of NaOH final is 0.05 / 0.250 = 0.2 mol dm^-3. The assumption is not volume = mass, but rather the volume of NaOH produced is small compared to 0.250 dm³ of water.