Hii there !
The reaction accompanying production of water in this case would be -

Thus ,by observing the reaction, we can conclude that 1 mole of Hydrogen (H
)react with 1/2 mole of oxygen(O
)to produce 1 mole of water (H
O)
Therefore, 37.0 moles of hydrogen(H
) will be required to produce 37.0 moles of water.
Hope it helps.
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 
<u>Explanation:</u>
We are given:
Initial moles of hydrogen sulfide gas = 0.47 moles
Volume of the container = 3.0 L
The molarity of solution is calculated by using the equation:

So, 
The given chemical equation follows:

<u>Initial:</u> 0.1567
<u>At eqllm:</u> 0.1567-2x 2x x
The expression of
for above equation follows:
![K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
We are given:

Putting values in above equation, we get:

So, equilibrium concentration of hydrogen gas = 
Hence, the concentration of hydrogen gas at equilibrium is 