A sample of 0.8360 g of an unknown compound containing barium ions (Ba2+) is dissolved in water and treated with an excess of Na2SO4. If the mass of the BaSO4 precipitate formed is 0.7794 g, what is the percent by mass of Ba in the original unknown compound?

percentage by mass of Ba = 0.4583/0.8360  × 100 = 45.83 /0.8360 = 54.82%

Explanation:

The mass of Barium in the original compound is the same as the masss of barium in the precipitate BaSO4.

Ba²+

molar mass of BaSO4 = 137 + 32 + 16(4) = 137 + 32 + 64 = 233 g/ mol

since

137 grams of barium is in 233 g of BaSO4

? grams of barium is in 0.7794 g of BaSO4

cross multiply

grams of barium =  0.7794 × 137/233

grams of barium = 106.7778/233

grams of barium = 0.45827381974

grams of barium = 0.4583 grams

percentage  by mass of Barium in the original compound  =

mass of barium/ mass of the compound × 100

mass of barium =  0.4583 g

mass of compound = 0.8360 g

percentage by mass of Ba = 0.4583/0.8360  × 100 = 45.83 /0.8360 = 54.82%

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.