We’re being asked to **calculate the rate for HF**** **for the following reaction:

6 NaOH + Al_{2}O_{3} + 12 HF → 2 Na_{3}AlF_{6} + 9 H_{2}O

Recall that for a reaction **aA ****→**** bB**, the ** rate of a reaction** is given by:

$\overline{){\mathbf{Rate}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{a}}\frac{\mathbf{\Delta}\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{\Delta t}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{b}}\frac{\mathbf{\Delta}\mathbf{\left[}\mathbf{B}\mathbf{\right]}}{\mathbf{\Delta t}}}$

where:

**Δ[A]** = change in concentration of reactants or products (in mol/L or M), *[A] _{final} – [A]_{initial}*

**Δt** = change in time, *t _{final} – t_{initial}*

Consider the following reaction:

6NaOH + Al_{2}O_{3} + 12HF → 2Na_{3}AlF_{6} + 9H_{2}O

At a given time, water is being produced at a rate of 0.54 M/min for this reaction. What is the corresponding rate for HF during this time?

A. -0.54 M/min

B. -0.72 M/min

C. -0.060 M/min

D. 0.54 M/min

E. 0.72 M/min

F. 0.060 M/min

G. -0.36 M/min

H. 0.36 M/min

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

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