Stoichiometry: moles/grams?

Question

A sample of 0.00329 moles of Na reacted according to the balanced equation:

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

After reaction, 5.70 x 10^-4 moles of Na remain. Calculate grams of NaOH formed.


i'm not sure how to set up the equation with the 0.00329 information. what do i do with 0.00329?

Answer 1

From the equation, it takes every 2 moles of Na to make 2 moles of NaOH, right? We have .00329 moles of Na from the start, and we have .00057 moles left. How much have we used?

.00329 - 0.00057 = .00272 mole have been used.

We know that there is 23g in 1 mol of Na, 16g in 1mol of O, and 1g in 1 mol of H. In summary, there are 40g in 1 mol of NaOH.
To get the mass in 0.00272 mol, we take 40g * 0.00272 mol equal 0.1088 grams of NaOH.

Ans. 40g*0.00272 = 0.1088 g NaOH

Answer 2

i'm uncertain if that's good via fact that we never did molecule to molecule stoichiometry, yet right here is what i could do. via fact the equation is already balanced, merely multiply all the coefficients by using 9. 9H2 + 9Cl2 ----> 18HCl in case you're at a loss for words and all, there is an "invisible" a million in front of the H2 and Cl2. We never placed a a million in front via fact a million x a style = the style. wish that enables!

Answer 3

moles=grams/M.W.
grams=moles*M.W.
(M.W. of NaOH is 40)