balanced equation: 2Na + 2H2O ---> 2NaOH + H2convert g of NaOH to moles using molar mass 34.5g NaOH x 1mol NaOH/40g NaOH = .8625 molthen find the mass of sodium metal that would be produced .8625 mol NaOH x 2 mol Na/2 mol NaOH x 23g Na/1mol Na = 19.8375 g Nadensity = mass/volume 0.97 g/mL = 19.......
15.6g Fe2O3 x 1mol Fe2O3/159.69g x 2mol Fe/1mol Fe2O3 x 55.845g Fe/1mol Fe= 10.91 g Fe
** the theoretical yield is 10.91g Fe **
The percent yield is found using (acutal mass produced/theoretical yield mass x 100)
(9.58g/10.91g) x 100 = 87.8
** the percent yield is 87.8% **
2. balanced equation: 2Na + 2H2O ---> 2NaOH + H2
convert g of NaOH to moles using molar mass
34.5g NaOH x 1mol NaOH/40g NaOH = .8625 mol
then find the mass of sodium metal that would be produced
.8625 mol NaOH x 2 mol Na/2 mol NaOH x 23g Na/1mol Na = 19.8375 g Na
density = mass/volume
0.97 g/mL = 19.8375 g / volume
volume = 20.451 mL
** You would need 20.45 mL of sodium metal **
Hope this helped :)
1. Start with balanced equation: Fe2O3 + 3CO ---------------> 2 Fe + 3CO2
2. Write what you are given below the equation.
Fe2O3 + 3CO ---------------> 2 Fe + 3CO2
m = 15.6 g m = 12.8 g
3. Convert all masses into moles
moles of Fe2O3 = 15.6 g / 159.7 g/mol = 0.09768 mol
moles of CO = 12.8 g / 32.01 g/mol = 0.3999 moles
4. Here is where it gets tricky. The equation says that for every mole of Fe2O3 used you need to have 3 times that many moles of CO. So you have 0.09768 mol of Fe2O3, so in order to use it all up you would need to have 3 times that much CO. 0.09768 mol x 3 = 0.293 moles. Whew!! you're safe, you have way more many moles than you need to use up all the Fe2O3. So the Fe2O3 is the limiting reagent.
5. OK. So 0.09768 mol of Fe2O3 is used. Now look at the mol ratio between Fe2O3 and Fe from the equation. Can you see that it is 1 to 2? OR, for every mole of Fe2O3 used you will produced 2 x as many moles of Fe
So moles of Fe will be 0.09768 mol x 2 = 0.19563 mol of Fe
6. Now convert that into a mass. Mass = moles x Molar mass
Mass (Fe) = 0.19563 mol x 55.85 g/mol = 10.91 g
That is the THEORETICAL yield. But you only got 9.58 g
7. Percent yield = (9.58 / 10.91) x 100 = 88%
OK?
As for your second question, I'm a little confused. Sodium metal doesn't come in milliliters, I'm not sure what you are asking for.
