Question

Define limiting reactant.20gram of 90% pure CaCO3 is allowed to react with 50gram 36.5% pure HCl by mass. Find the
a.Limiting reactant
b. Number of mole of excess reactant left over unreacted
c.Mass of the salt formed
d. Volume of CO2 produced
​

Answer 1

Answer:In a chemical reaction, the limiting reagent is the reactant that determines the quantity of the products that are produced. The other reactants present in the reactions are sometimes found to be in excess since there is some leftover quantity of them after the limiting reagent is completely used up.

Explanation:

By this way

Answer 2

Answer:

CaCO₃

0.14

19.98g

4.45ml

Explanation:

As per the data given,

We have,

20g of 90% pure CaCO₃,

Reacting with 50g of 36.5% pure HCl

Reaction is :

CaCO₃ + 2HCl →CaCl₂ + CO₂ + H₂O

Molar mass CaCO₃ : 100

Molar mass HCl : 36.5

20g of 90% of CaCO₃ = 20*90/100= 18g

50g of 36.5% of HCl= 50*36.5/100=18.25g

100 g of CaCO₃ uses  2* 36.5g of HCl

so,  18g of CaCO₃ uses  (2* 36.5/100)*18g HCl

= 13.14g HCl

Similarly,

36.5 g HCl →1/ 2*100g CaCO₃

18.25g HCl → (100/36.5*2)*18.25g CaCO₃

= 25g CaCO₃

But we only have 18g of CaCO₃

And as we know a limiting reactant is the one that finishes early during the reaction.

∴CaCO₃ is the limiting reactant.

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

Limiting reactant : CaCO₃ = 18g

HCl = 18.25g

∴18g of CaCO₃ uses → (2* 36.5/100)*18g HCl

= 13.14g HCl

HCl used: 18.25g – 13.14 g= 5.11g

Moles = 5.11/36.5 = 0.14 mol

Molar mass of CaCl₂ = 111g

1 mole CaCO₃→ 1 mole CaCl₂

∴ 18/100 mol CaCO₃→ 18/100*111g of CaCl₂

=19.98g

Volume of CO₂ produced = moles*24.7

=18/100*24.7 = 4.45ml

(I mole CaCO₃ →1 mole CO₂ )

#SPJ2