# Problem: Consider the following reaction:6NaOH + Al2O3 + 12HF → 2Na3AlF6 + 9H2OAt 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/minB. -0.72 M/minC. -0.060 M/minD. 0.54 M/minE. 0.72 M/minF. 0.060 M/minG. -0.36 M/minH. 0.36 M/min

###### FREE Expert Solution

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

6 NaOH + Al2O3 + 12 HF → 2 Na3AlF6 + 9 H2O

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, tfinal – tinitial

###### Problem Details

Consider the following reaction:

6NaOH + Al2O3 + 12HF → 2Na3AlF6 + 9H2O

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