Consider the reaction, 2 d(g) + 3 e(g) + f(g) => 2 g(g) + h(g) when h is increasing at 0.64 mol/ls, how quickly is e decreasi
ng? give your answer to 3 decimal places.
1 answer:
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.
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