13 Calcium carbonate reacts with hydrochloric acid: CaCO_(3)(s)+2HCl(aq)arrow CaCl_(2)(aq)+H_(2)O(l)+CO_(2)(g) Calcium carbonate is added to 25.0cm^3 of 2.00mol/dm^3 hydrochloric acid. (A_(r):C=12,O=16,Ca=40) Calculate the amount in moles of hydrochloric acid. Calculate the amount in moles of CaCO_(3) that reacts with the acid. Calculate the mass of CaCO_(3) that reacts with the acid. Calculate the volume of CO_(2) (measured at rtp)produced. (Molar volume at rtp is 24000cm^3

Question

Answer 1

Let's go through each part of the problem step by step:### 1. Calculate the amount in moles of hydrochloric acid.Given:- Volume of hydrochloric acid,

- Concentration of hydrochloric acid,

First, convert the volume from cm³ to dm³:

Now, calculate the amount in moles:

### 2. Calculate the amount in moles of

that reacts with the acid.From the balanced chemical equation:

The stoichiometry shows that 1 mole of

reacts with 2 moles of HCl.Given that we have 0.050 moles of HCl, the moles of

that can react are:

### 3. Calculate the mass of

that reacts with the acid.Given:- Molar mass of

: -

-

Now, calculate the mass:

### 4. Calculate the volume of

produced (measured at rtp).From the balanced chemical equation, 1 mole of

produces 1 mole of

.Given that we have 0.025 moles of

, the moles of

produced are:

Given:- Molar volume of

at rtp = 24000 cm³/molNow, calculate the volume:

### Summary of Results:1. Amount in moles of hydrochloric acid:

2. Amount in moles of

that reacts with the acid:

3. Mass of

that reacts with the acid:

4. Volume of

produced:

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