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Nata [24]
3 years ago
11

A 211 g sample of barium carbonate, baco3, reacts with a solution of nitric acid to give barium nitrate, carbon dioxide, and wat

er. if the acid is present in excess, what mass and volume of dry carbon dioxide gas at stp will be produced?​
Chemistry
1 answer:
bija089 [108]3 years ago
8 0
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
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3 years ago
Consider the following data:
VikaD [51]

∆H ° rxn =-2855.56 kJ

<h3>Further explanation</h3>

Given

ΔHf CO₂ = -393.5 kJ/mol

ΔHf H₂O = -241.82 kJ/mol

ΔHf  C₂H₆ = - 84.68 kJ/mol

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2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)

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ΔHrxn=

Solution

<em>∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants) </em>

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∆H ° rxn = (-1574-1450.92)-(-169.36)

∆H ° rxn =-3024.92+169.36

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