Barium can be analyzed by precipitating it as bas04 and weighing the precipitate. when a 0.713-g sample of a barium compound was treated with excess h2s04, 0.5331 g of baso4 formed. what is the percentage of barium in the compound?
molecular mass of baso4 233.39 g/mol
molecular mass of ba = 137.327 g/mol

31.37%

Explanation:

For this case, you should consider the following reaction:

Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O

For which you obtain the precipitate of BaSO₄

In order to obtain the mas of Barium on the precipitate, you may use the following formula:

gBa= M₍BaSO₄₎x(M₍Ba₎/M₍BaSO₄₎)

Where:

gBa= mass of Barium

M₍BaSO₄ ₎= mass of BaSO₄ from the precipitate

M₍Ba₎= mass of Barium from the original sample

M₍BaSO₄₎= mass of BaSO₄ from the precipitate

gBa= (0.5331)x(137.327/233.39)= 0.3136 g

Then we ontain the percentage of Barium multiplying by 100:

% Ba on the original sample= 31.36%

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: