Which of the following `is//are` correct?
The following reaction occurs:
`Na_(2) CO_(3) + 2HCl rarr 2NACl + CO_(2) + H_(2) O`
`106.g "of" Na_(2) CO_(3)` reacts with `109.5 g "of" HCl`.

Options

A. The `HCl` is in excess.
B. `117.0 g` of `NaCl` is formed.
C. The volume of `CO_(2)` produced at 1 bar and `273 K` is `22.7 L`
D. The volume of `CO_(2)` produced at 1 bar and `298 K` is `24.7 L`

Answer

A, B, C, D

Explanation

`(Mw "of" Na_(2) CO_(3) = 106 "of" HCl = 36.5, Mw "of" NaCl = 58.5)`
Mole of `Na_(2) CO_(3) = (106)/(106) = 1.0 "mol"`
Moles of `HCl = (109.5)/(36.5) = 3.0 "mol"`
a. Since for a mol of `Na_(2) CO_(3)`, 2 mol of `HCl` is required.
So, `HCl` is in excess `(3 - 2) = 1.0 "mol"`
Therefore, `Na_(2) CO_(3)` is the limiting quaintity.
b. weight of `NaCl` formed
`= (1.0 "mol" Na_(2) CO_(3)) ((2 "mol" NaCl)/("mol" Na_(2) CO_(3)))((58.8 g NaCl)/("mol" NaCl))`
1 mol of `Na_(2) CO_(3) = 1` mol of `CO_(2) = 22.7 L` at 1 bar,
`273 K`
1 mol of `Na_(2) CO_(3) = 1` mol of `CO_(2) = 24.7 L` at 1 bar,
`298 K`