Question #f5f0c
2 Answers
Explanation:
The trick here is to realize that after the initial sample is heated, you will only be left with strontium oxide,
Heating the sample will consume all the strontium carbonate,
In other words, the difference between the initial mass of the sample and the final mass of the is equal to the mass of carbon dioxide evolved by the reaction.
You can thus say that the reaction produced
#overbrace("1.850 g")^(color(blue)("initial mass")) - overbrace("1.445 g")^(color(purple)("final mass")) = "0.405 g"#
of carbon dioxide. Use the compound's molar mass to convert this to moles
#0.405 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.009202 moles CO"_2#
Now, write down the balanced chemical equation that describes this decomposition reaction
#"SrCO"_ (3(s)) stackrel(color(red)(Delta)color(white)(aa))(->) "SrO"_ ((s)) + "CO"_(2(g))# #uarr#
Notice that every mole of strontium carbonate that undergoes decomposition produces
#0.009202 color(red)(cancel(color(black)("moles CO"_2))) * "1 mole SrCO"_3/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.009202 moles SrCO"_3#
Next, use the compound's molar mass to convert the number of moles to grams
#0.009202 color(red)(cancel(color(black)("moles SrCO"_3))) * "147.63 g"/(1color(red)(cancel(color(black)("mole SrCO"_3)))) = "1.3585 g"#
Finally, you can say that the percent composition of strontium carbonate in the original sample was
#"% SrCO"_3 = (1.3585 color(red)(cancel(color(black)("g"))))/(1.850color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(73.43%)color(white)(a/a)|)))#
The answer is rounded to four sig figs.
73.41%
Explanation:
The amount of
We then use the molecular weights of Sr
Mass of
0.405g/44.01 g/mol
Because the ratio of
The final desired mass ratio is then 1.358/1.850 = 0.7341,
or 73.41% Sr