# Question 64d6d

May 8, 2016

${\text{2.47 moles SrCl}}_{2}$

#### Explanation:

Since you didn't provide a chemical equation that describes the reaction you're interested in, I will assume that you're dealing with this one

${\text{SrO"_ ((s)) + 2"HCl"_ ((aq)) -> "SrCl"_ (2(aq)) + "H"_ 2"O}}_{\left(l\right)}$

Strontium oxide, $\text{SrO}$, will react with hydrochloric acid, $\text{HCl}$, to form strontium chloride, ${\text{SrCl}}_{2}$, and water.

No mention was made of the amount of hydrochloric acid you have at your disposal, so you can safely assume that it's in excess. This means that all the moles of strontium oxide present in that $\text{256-g}$ sample will take part in the reaction.

Use strontium oxide's molar mass to determine how many moles of each you have present

256 color(red)(cancel(color(black)("g"))) * "1 mole SrO"/(103.62color(red)(cancel(color(black)("g")))) = "2.47 moles SrO"#

Now, the reaction will consume strontium oxide and produce strontium chloride in a $1 : 1$ mole ratio. This means that you will have

$2.47 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles SrO"))) * "1 mole SrCl"_2/(1color(red)(cancel(color(black)("mole SrO")))) = color(green)(|bar(ul(color(white)(a/a)"2.47 moles SrCl}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.