Question

The volume of 3M HCL required to completely react with 1.431g of sodium carbonate? pls answer the question I will follow u and give you brainly if it is correct

a)10 ml
b)9ml
c)8ml
d)7ml​

Answer 1

Answer:

9 ml

Explanation:

To answer this question you first write the correct chemical equation.

2 HCl + Na2Co3 —>2NaCl + CO2 + H2O

We then proceed to solve it by finding the moles of sodium carbonate

to find moles we take the mass divide by the molar mass

Molar mass

Na = 23

C = 12

O = 16

Therefore

23 × 2 = 46

C = 12

O = 16 × 3 = 48

We then add all of the to get the molar mass of Sodium carbonate

46 + 12 + 48 = 106

We now proceed to getiing the moles

1.431 g ÷ 106 = 0.0135 moles

The mole ratio of HCl to Sodium carbonate is 2:1

Therefore we take

0.0135 × 2 = 0.027 moles of HCl

0.027 × 1000 =27

We then proceed to divide

27÷3 = 9 cm3

1cm3 is equivalent to 1 ml

Hence 9 ml

Answer 2

Explanation:

here's your answer

9ml

hope you GET it