Answer:
Mass = 47.04 g
Volume = 23.94 L
Solution:
The equation for given reaction is as follow,
BaCO₃ + 2 HNO₃ → Ba(NO₃)₂ + CO₂ + H₂O
According to this equation,
197.34 g (1 mole) BaCO₃ produces = 44 g (1 mole) of CO₂
So,
211 g of BaCO₃ will produce = X g of CO₂
Solving for X,
X = (211 g × 44 g) ÷ 197.34 g
X = 47.04 g of CO₂
As we know,
44 g (1 mole) CO₂ at STP occupies = 22.4 L volume
So,
47.04 g of CO₂ will occupy = X L of Volume
Solving for X,
X = (47.04 g × 22.4 L) ÷ 44 g
X = 23.94 L Volume
Answer:
There's no picture, so I can't help witht his, apologies!
Explanation:
Answer:
C21H2802
Explanation:
C=12g/mol
H=1g/mol
O=16g/mol
Part (C) of compound-80.18%
(0.8018 x 312)/12=21
Part (H) of compound-8.97%
(0.897 x 312)/1=28
Part (O) of compund-10.3%
(0.103 x 312)/16 = 2
Therefore the emp. formula is C12H28O2
