Answer:
10.80
Explanation:
As per the equation, let us calculate the mole ratio. N2+3H2→2NH3. As per the equation one mole of nitrogen reacts with 1 mol of hydrogen.
In terms of mass. 28.01 g of nitrogen needs 3 mol of hydrogen or 6.048 g of hydrogen.
We can set up the ratio;
28.01 g of
l
N
2
needs
6.048 g of
l
H
2
1 g of
l
N
2
needs
6.048
28.01
g of
l
H
2
50.0 g of
l
N
2
needs
6.048
×
50.0
28.01
l
g of
l
H
2
=
10.80 g of
l
H
2
Answer:
Extensive is the answer of this question?
D) Chlorine is reduced as hydrogen is added whereas Iron is oxidised as Joins with a non hydrogen compound.
Since Chlorine is in excess, this is a limiting reagent problem.
1) convert 11.50 g Na to g of NaCl using the balanced equation.
2) percent yield = (actual yield)/(potential yield).
.85= (actual yield)/(g from step 1)
Solve for actual yield
Answer:
Concentration of original solution = 1.66
Explanation:
We know that
We have given concentration of NaOH = 0.1678
Volume of NaOH = 19.88 mL = 0.01988 L
So moles of NaOH = volume x concentration of NaOH
= 
Moles of 
in 10 mL of diluted solution = 1/2 x moles of NaOH
= 
x 0.00333 = 0.00166 mol
Moles of in 25 mL of original solution
= moles of H2SO4 in 250 mL of diluted solution
= 
x 0.00166 = 0.0415 mol
Concentration of original solution = 
= 
