Question #eac5b
1 Answer
Explanation:
Your tool of choice here will be the ideal gas law equation
color(blue)(PV = nRT)" "PV=nRT , where
As you can see, this form of ideal gas law equation does not allow you to calculate the mass of the gas directly. However, you can use the molar mass of the gas to find a relationship between the mass of a sample and the number of moles it contains.
As you know, the molar mass of a substance tells you the mass of one mole of that substance.
color(blue)(M_M = m/n)" "MM=mn , where
Notice that you can use this equation to find the number of moles
M_M = m/n implies n = m/M_MMM=mn⇒n=mMM
Plug this into the ideal gas law equation to get
PV = m/M_M * RTPV=mMM⋅RT
Rearrange to solve for
PV * M_M = m * RT implies m = M_M * (PV)/(RT)PV⋅MM=m⋅RT⇒m=MM⋅PVRT
Now, before you plug in your values, make sure that the units you're working with match those used by the universal gas constant.
{:(color(red)("Need"), color(white)(aaaaa), color(blue)("Have")), (color(white)(aaaa), color(white)(aaaa), color(white)(aaaa)), (color(white)(aa)"L", color(white)(aaaa), color(white)(a)"mL"color(white)(aaaaaa)), ("atm", color(white)(aaaa), color(white)(a)"torr"color(white)(aaaaa)), (color(white)(aa)"K", color(white)(aaaa), color(white)(a)""^@"C"color(white)(aaaaaa)) :}
You will thus have to use the following conversion factors
"1 atm " = " 760 torr"
"1 L" = 10^3"mL"
T = t^@"C" + 273.15
Carbon dioxide,
m = 44.01"g"/color(red)(cancel(color(black)("mol"))) * ( 625/760color(red)(cancel(color(black)("atm"))) * 27 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))
m = "0.03992 g"
Rounded to two sig figs, the answer will be
m = color(green)("0.40 g CO"_2)
