Chemistry molecular formular?

Question

When Barium Carbonate is heated strongly, it decomposes to give carbon dioxide and barium oxide.

BaCO3(s) --> BaO(s) + CO2 (g)

What is the volume of carbon dioxide produced at r.t.p when 78.8g of barium carbonate is heated?
[Relative molecular mass of barium carbonate is 197]

how do i convert the answer to dm3 then ?

Answer 1

First find the moles of BaCO3:

n = m/Mr = 78.8/197 = 0.4 mol, where Mr = 197 the relative molar mass of BaCO3.

Then from the balanced chem. equation: BaCO3 --> BaO + CO2

1 mol of BaCO3 gives 1 mol of CO2, so
0.4 mol of BaCO3 gives 0.4 mol of CO2

At last find the volume of CO2 at S.T.P.

V = n*Vm = 0.4*22.4 = 8.96 L, where Vm = 22.4 L/mol the molar volume of a gas at S.T.P. (1 atm, 0 oC)

Answer 2

one BaCO3 forms one BaO and one CO2
197 gives 38
78.8 gives 78.8 *38 /197
i don't have a calc. oops, forgot the one on the comp
15.2 now, this is true if the equation is balanced and rtp means something like regular temperature and pressure. i was taught stp, standart temp and pressure.
hope i got it right

Answer 3

197g BaCO3 gives 12+32=44g CO2
at stp, 44g CO2 = 22.4 L
197g BaCO3 gives 22.4 L
78.8g BaCO3 gives 22.4 / 197 * 78.8 CO2 = 8.96 L