What mass of NaOH is produced when 0.25 g of Na reacts with excess H2O in the following reaction?

Question

What mass of NaOH is produced when 0.25 g of Na reacts with excess H2O in the following reaction? 2 Na + 2 H2O ® 2 NaOH + H2

Answer 1

2 Na + 2 H2O ----> 2 NaOH + H2

you'll need to convert the 0.25 grams of Na into moles:

0.25 g Na x [1 mol Na / 23 g Na] = 0.01 mol Na

now use the coefficients on the balanced equation to convert the 0.01 mol Na into moles of NaOH:

0.01 mol Na x [2 mol NaOH / 2 mol Na] = 0.01 mol NaOH

the last step is to convert the 0.01 mol NaOH into grams:

0.01 mol NaOH x [40 g NaOH / 1 mol NaOH] = 0.4 g NaOH

Answer 2

Divide 0.25 g by g-atom weight of sodium, get g-atoms of Na+ (call this A)
Find mole wt of NaOH
Multiply mole wt of NaOH by A

Answer 3

Calculate the variety of moles of Li3N you initiate up with; the balanced equation tells you which you get 3 moles of LiOH for each mole of Li3N you started out with. So...in case you initiate up with a million mole of Li3N you will get 3 moles of LiOH; 2 moles of Li3N, 6 moles of LiOH; 0.5 moles of Li3N, a million.5 moles of LiOH, and so on. After determining how many moles of LiOH you get, you ought to transform that to grams.