Answer:
304.19 g
Explanation:
M(H2) = 2*1 = 2 g/mol
54.1 g H2 * 1 mol H2/2gH2 = 27.05 mol H2
3H2 + N2 ------> 2NH3
from reaction 3 mol 2 mol
given 27.05 mol x mol
x = 27.05*2/3 = 18.03 mol NH3
M(NH3) = 14 +3*1 = 17 g/mol
18.03 mol NH3 * 17 g NH3/ 1 mol NH3 ≈ 307 g
Closeet answer is 304.19 g.
<span>we are going to use H-H equation:
PH = Pka + log[conjugate base]/[weak acid]
when we have here the conjugate base is NaF
and the weak acid is HF.
so, we are going to use the Ka value to get the value of Pka.
when Pka = -logKa
= -log7.2 X 10^-4
= 3.14
then we will substitution on H-H equation:
</span>∴<span>PH = 3.14 + log (0.623)/(1.41) = 2.99</span>
Explanation:
Aluminium - Al
Iron Sulfate - FeSO4
The reaction is given as;
Al + FeSO₄ --> Al₂ (SO₄) ₃ + Fe
The balanced equation is given as;
2 Al + 3 FeSO₄ --> Al₂ (SO₄) ₃ + 3 Fe
In words it is given as;
Aluminium + Iron Sulfate --> Aluminium Sulfate + Iron
