Question

When 4 g of a metal carbonate MCO, was dissolved in 160 cm of 1 M hydrochloric acid and then the resultant solution diluted to one litre, 25. 0 cm of this solution required 20. 0 cm' of 0. IM sodium hydroxide solutidn for complete neutralisation, calculate:

(i) The number of moles per litre of excess hydrochloric acid that reacted with sodium hydroxide, NaOH.

(ii) The number of moles per litre of acid that reacted with the carbonate.

(iii) The number of moles of carbonate, MCO, that reacted with the acid.

(iv) The formula mass of the carbonate, MCO, (v) The atomic mass of the metal M. (C = 12. 0. 0 = 16. 0)​

Answer 1

(i) The number of moles per litre of excess hydrochloric acid that reacted with sodium hydroxide, NaOH, is 0.008 mol/L.

(ii) The number of moles per litre of acid that reacted with the carbonate is 0.025 mol/L.

(iii) The number of moles of carbonate, MCO, that reacted with the acid is 0.004 mol.

(iv) The formula mass of the carbonate, MCO, is 88 g/mol.

(v) The atomic mass of the metal M is 24 g/mol.

Step-by-step explanation:

(i) The reaction between HCl and NaOH gives water and NaCl. Using the equation:

HCl + NaOH → NaCl + H2O

Given that 25.0 cm³ of the solution reacted with 20.0 cm³ of 0.1M NaOH, we can find the moles of NaOH used:

0.1 mol/L × 20.0 cm³ ÷ 1000 = 0.002 mol.

Since HCl and NaOH react in a 1:1 ratio, the moles of HCl that reacted with NaOH are also 0.002 mol/L.

(ii) The excess HCl that remains can be calculated. Initially, 4g of MCO reacted with HCl to form CO2, H2O, and MCl2. From the volume of HCl used, the moles of HCl reacted with the carbonate is:

1 mol of HCl reacts with 1 mol of MCO, hence 0.004 mol of HCl reacted with MCO.

So, in 1 L of solution, the moles of HCl that reacted with the carbonate = 0.004 mol.

(iii) To find the moles of carbonate that reacted with the acid, it's derived from the stoichiometry of the balanced equation, indicating that 1 mol of MCO reacts with 2 moles of HCl. Therefore, 0.004 mol of HCl reacted with MCO, indicating 0.002 mol of MCO reacted with 0.004 mol of HCl.

(iv) The formula mass of MCO is found by adding the atomic masses of the constituent atoms: M (metal) + C + O × 3 = 24 + 12 + 16 × 3 = 88 g/mol.

(v) The atomic mass of the metal M can be calculated by subtracting the known atomic masses of C and O from the total formula mass of MCO: 88 - 12 - 16 × 3 = 24 g/mol.

Answer 2

To solve this question, first, we determine the number of moles per litre of excess hydrochloric acid that reacted with sodium hydroxide. Then, we find the number of moles per litre of acid that reacted with the carbonate. Next, we calculate the number of moles of carbonate that reacted with the acid. After that, we determine the formula mass of the carbonate and finally, find the atomic mass of the metal M.

Step-by-step explanation:

To solve this question, we need to determine the number of moles per litre of excess hydrochloric acid that reacted with sodium hydroxide (NaOH), the number of moles per litre of acid that reacted with the carbonate (MCO), the number of moles of carbonate (MCO) that reacted with the acid, the formula mass of the carbonate (MCO), and the atomic mass of the metal M. Let's break it down step-by-step:

1. Using the balanced chemical equation, we know that there is a one-to-one ratio of moles of HCl to moles of NaOH. Therefore, the number of moles per litre of excess hydrochloric acid that reacted with sodium hydroxide is equal to the number of moles per litre of sodium hydroxide, which is 0.0200 mol/L.
2. Since the number of moles per litre of acid that reacted with the carbonate is equal to the number of moles per litre of excess hydrochloric acid that reacted with sodium hydroxide, the answer is also 0.0200 mol/L.
3. The number of moles of carbonate (MCO) that reacted with the acid can be calculated by multiplying the number of moles per litre of acid that reacted with the carbonate and the volume of acid used, which is 0.0200 mol/L * 0.025 L = 0.0005 mol.
4. To find the formula mass of the carbonate (MCO), we need to determine the molar mass of MCO. From the equation MCO, we know that it contains one atom of metal M, one atom of carbon (C), and three atoms of oxygen (O). The molar mass of MCO is therefore equal to the atomic mass of M plus the atomic mass of C plus three times the atomic mass of O, which is (atomic mass of M) + 12.01 + (3 * 16.00).
5. Finally, to find the atomic mass of the metal M, we subtract the atomic mass of C and three times the atomic mass of O from the formula mass of the carbonate (MCO).