# Question 0b1c2

May 25, 2017

Here's what I got.

#### Explanation:

You know that the following reaction

${\text{CaCO"_ (3(s)) + color(red)(2)"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

has an initial rate in terms of the mass of carbon dioxide produced per second

${\text{rate" = "2.0 g CO"_2color(white)(.)"s}}^{- 1}$

The first thing that you need to do here is to convert this to number of moles of carbon dioxide per second by using the compound's molar mass

2.0 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.04544 moles CO"_2

You can thus say that the initial rate of the reaction is

${\text{rate" = "0.04544 moles CO"_2color(white)(.)"s}}^{- 1}$

Now, to convert this to moles of hydrochloric acid per hour, use the $\textcolor{red}{2} : 1$ mole ratio that exists between the two compounds and the fact that

$\text{1 hr = 3600 s}$

Do not forget that you need to add a minus sign because the concentration of hydrochloric acid is decreasing as the reaction proceeds!

You should end up with

$\text{rate" = - (0.04544 color(red)(cancel(color(black)("moles CO"_2))))/(1 color(red)(cancel(color(black)("s")))) * (color(red)(2)color(white)(.)"moles HCl")/(1color(red)(cancel(color(black)("mole CO"_2)))) * (3600 color(red)(cancel(color(black)("s"))))/"1 hr}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{rate" = -"330 moles HCl"color(white)(.)"hr}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the ${\text{2.0 g CO}}_{2}$ ${\text{s}}^{- 1}$ rate.

To convert the initial rate to liters of carbon dioxide per minute at STP, use the fact that the molar volume of a gas at STP is usually given as being equal to $\text{22.4 L}$ and the fact that

$\text{1 min = 60 s}$

You don't need a minus sign here because the concentration of carbon dioxide is increasing as the reaction proceeds.

You should end up with

$\text{rate" = (0.04544 color(red)(cancel(color(black)("moles CO"_2))))/(1 color(red)(cancel(color(black)("s")))) * "22.4 L"/(1color(red)(cancel(color(black)("mole CO"_2)))) * (60 color(red)(cancel(color(black)("s"))))/"1 min}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{rate" = "61 L CO"_2color(white)(.)"min}}^{- 1}}}}$

Finally, to convert the initial rate to grams of calcium carbonate per hour, start by converting the number of moles of carbon dioxide to moles of calcium carbonate

0.04544 color(red)(cancel(color(black)("moles CO"_2))) * "1 mole CaCO"_3/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.04544 moles CaCO"_3

Convert this to grams by using the compound's molar mass

0.04544 color(red)(cancel(color(black)("moles CaCO"_3))) * "100.9 g"/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "4.585 g"#

This means that you have

${\text{2.0 moles CO"_2color(white)(.)"s"^(-1) = "4.585 g CaCO"_3color(white)(.)"s}}^{- 1}$

Once again, the amount of calcium carbonate decreases as the reaction proceeds, so don't forget about the minus sign!

You should end up with

$\text{rate" = - "4.585 g CaCO"_3/(1 color(red)(cancel(color(black)("s")))) * (3600 color(red)(cancel(color(black)("s"))))/"1 hr}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{rate" = - "17000 g CaCO"_3color(white)(.)"hr}}^{- 1}}}}$

Once again, the answer is rounded to two sig figs.