4. Calculate the mass of zinc carbonate that would remain if 17.0g of zinc carbonate was
reacted with 50cm3 of 4M nitric acid. The equation of the reaction is:
ZnCO3 (g) + 2HNO3 (aq) Zn (NO3) 2 + CO2 (g) + H2O (l)
(Zn=65.4, C=12.0, O=16.0)
A certain carbonate XCO3 , reacts with dilute hydrochloric acid according to the
equation given below:
XCO3(s) +2HCl (aq) XCl2 (aq) + CO2 (g) + H2O (l)
ZnCO3 (g) + 2HNO3 (aq) Zn (NO3) 2 + CO2 (g) + H2O (l)
millimoles HNO3 = mL x M = 50 x 4 = 200
That 200 mmols HNO3 will use 1/2 x 200 = 100 mmols Zn or 0.1 mols Zn.
How much Zn did you have? That's mols Zn = g/atomic mass = 17.0/65.4 0.26. You used 0.1. Remaining Zn = 0.26- 0.1 = 0.16 mols. grams Zn = 0.16 mols x 65.4 = ?
To calculate the mass of zinc carbonate that would remain if 17.0g of zinc carbonate was reacted with 50cm3 of 4M nitric acid, we need to first determine the limiting reactant.
Step 1: Calculate the number of moles of zinc carbonate:
Number of moles = Mass / molar mass
Molar mass of ZnCO3 = (Zn + C + 3O) = (65.4 + 12.0 + 3(16.0)) = 125.4 g/mol
Number of moles of ZnCO3 = 17.0g / 125.4 g/mol = 0.136 mol
Step 2: Calculate the number of moles of nitric acid:
Number of moles = Volume (L) * Concentration (M)
Volume in L = 50cm3 = 50/1000 = 0.05 L
Number of moles of nitric acid = 0.05 L * 4 M = 0.2 mol
Step 3: Determine the limiting reactant:
The balanced equation tells us that 1 mole of ZnCO3 reacts with 2 moles of HNO3. Therefore, the ratio of ZnCO3 to HNO3 is 1:2.
Since we have 0.136 mol of ZnCO3 and 0.2 mol of HNO3, we can see that the number of moles of ZnCO3 is less than half of the number of moles of HNO3. Thus, ZnCO3 is the limiting reactant.
Step 4: Calculate the number of moles of zinc nitrate produced:
From the balanced equation, we can see that the ratio of ZnCO3 to Zn(NO3)2 is 1:1.
Therefore, the number of moles of Zn(NO3)2 produced is equal to the number of moles of ZnCO3 reacted, which is 0.136 mol.
Step 5: Calculate the mass of zinc nitrate:
Mass = Number of moles * Molar mass
Molar mass of Zn(NO3)2 = (Zn + 2(NO3)) = (65.4 + 2(14.0 + 3(16.0))) = 189.4 g/mol
Mass of zinc nitrate = 0.136 mol * 189.4 g/mol = 25.78 g
Therefore, the mass of zinc carbonate that would remain if 17.0g of zinc carbonate was reacted with 50cm3 of 4M nitric acid is 17.0g - 25.78g = -8.78g. However, the mass cannot be negative. Thus, there would be no zinc carbonate remaining after the reaction.
To calculate the mass of zinc carbonate that would remain after the reaction, we need to use stoichiometry and the concept of limiting reactant.
Step 1: Calculate the number of moles of zinc carbonate (ZnCO3) and nitric acid (HNO3) using their respective masses and molar masses.
Given:
Mass of ZnCO3 = 17.0 g
Molar mass of ZnCO3 = (Zn: 65.4 g/mol) + (C: 12.0 g/mol) + 3(O: 16.0 g/mol) = 125.4 g/mol
Number of moles of ZnCO3 = Mass of ZnCO3 / Molar mass of ZnCO3
= 17.0 g / 125.4 g/mol
Step 2: Calculate the number of moles of nitric acid (HNO3) from the volume and molarity.
Given:
Volume of HNO3 = 50 cm³ = 50 * 10⁻³ L
Molarity of HNO3 = 4 M (mol/L)
Number of moles of HNO3 = Volume of HNO3 * Molarity of HNO3
= 50 * 10⁻³ L * 4 mol/L
Step 3: Use the stoichiometry of the balanced equation to determine the mole ratio between ZnCO3 and HNO3.
From the balanced equation:
1 mole of ZnCO3 reacts with 2 moles of HNO3
Step 4: Identify the limiting reactant, which is the reactant that is completely consumed first and determines the amount of product formed.
To determine the limiting reactant, compare the number of moles of ZnCO3 and HNO3 calculated in Step 1 and Step 2, respectively. The reactant with the smaller number of moles is the limiting reactant.
Step 5: Calculate the number of moles of ZnCO3 that react with HNO3.
Based on the stoichiometry of the balanced equation, we can determine that the number of moles of ZnCO3 that react with HNO3 is (1/2) times the number of moles of HNO3.
Number of moles of ZnCO3 reacting with HNO3 = (1/2) * Number of moles of HNO3
Step 6: Calculate the mass of the remaining zinc carbonate.
The mass of the remaining zinc carbonate can be calculated using the number of moles of ZnCO3 remaining (number of moles of ZnCO3 initially - number of moles of ZnCO3 that react with HNO3) and the molar mass of ZnCO3.
Mass of remaining ZnCO3 = Number of moles of ZnCO3 remaining * Molar mass of ZnCO3
That's how you can calculate the mass of zinc carbonate that would remain if 17.0g of zinc carbonate was reacted with 50cm³ of 4M nitric acid.