Question #4492b
1 Answer
Explanation:
The idea here is that you need to use the balanced chemical equation that describes this single replacement reaction to determine how many grams of hydrochloric acid,
#"Mg"_ ((s)) + color(red)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + "H"_(2(g)) uarr#
Notice that you have a
Use hydrogen gas' molar mass to determine how many moles you have in that
#10 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "4.96 moles H"_2#
This means that the reaction must have consumed
#4.96 color(red)(cancel(color(black)("moles H"_2))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole H"_2)))) = "9.92 moles HCl"#
Now use hydrochloric acid's molar mass to determine how many grams would contain this many moles
#9.92 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = "361.7 g HCl"#
Now, you know that you're working with a solution that
As it turns out, you don't need to use the density of the solution to find the volume that would contain
So, if get
#361.7 color(red)(cancel(color(black)("g HCl"))) * overbrace("100 mL solution"/(27color(red)(cancel(color(black)("g HCl")))))^(color(purple)("= 27% w/v")) = "1340 mL"#
of solution to provide
I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig figf for the mass of hydrogen gas produced by the reaction
#"volume of 27% w/v HCl solution" = color(green)(|bar(ul(color(white)(a/a)"1300 mL"color(white)(a/a)|)))#