Answer is: 1,92 mol/L·s.
Chemical reaction: 2D(g) + 3E(g) + F(g) → <span>2G(g) + H(g).
</span>H is increasing at 0,64 mol/L·<span>s.
From chemical reaction n(H) : n(E) = 1 : 3.
0,64 mol : n(E) = 1 : 3.
n(E) = 1,92 mol.
</span>E is decreasing at 1,92 mol/L·s.
The grams of CuCl2 required to prepare 1250g of 1.25% by mass is calculated as follow
1250 = 100%
what about 1.25%
by cross multiplication
= 1.25 x1250/100 = 15.63 grams
