FeS2 + 5 O2 2Fe2 + 4SO2 How many moles of FeO are formed from 0.57 moles of O2?

Chemistry

1 answer:

bagirrra123 [75]3 years ago

3 0

Answer:

0.228 mol FeO

Explanation:

2FeS2 + 5 O2----> 2FeO + 4SO2

from reaction 5 mol 2 mol

given 0.57 mol x

x = 0.57*2/5 = 0.228 mol FeO

You might be interested in

A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity

grandymaker [24]

Answer:

Mass of Ca in sample, Mass of Br in sample, Number of moles of Ca in sample, Number of moles of Br in sample, Mass or moles of element other than Ca or Br in sample

Explanation:

The AP Classroom will not count your answer to this question as correct unless it includes at least one of the answers listed above. If you say that theanswer to this question is density, it will be marked as incorrect, I found that out the hard way when I used the answers that brainly gave me.

Good luck,

I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.

8 0

2 years ago

Reading the above graph to the nearest tenth of a milliliter, what is the volume of sodium hydroxide at the equivalence point?

Vlada [557]

Answer:

The volume of sodium hydroxide at the equivalence point is:

14.9 mL of sodium hydroxide.

Explanation:

The equivalence point occurs when, in this case, the HCl is completely neutralized with the solution of NaOH, how you can see this doesn't occur in the last point but occurs in the nineteenth point, where the pH is no more acid (below to 7) but is 11 approximately, then you must see in the X-axis from this point and you can see the volume is almost 15, by this reason I calculate the valor of 14.9 milliliters.

7 0

3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$