Theoretically how many grams of malachite should be formed when 1.5 moles of Copper Sulphate reacts with an excess of sodium carbonate?
165.837
CuSO4 + Na2CO3 ==> CuCO3 + Na2SO4
mols CuSO4 = 1.5
mols CuCO3 = 1.5 x (1 mol CuCO3/1 mol CuSO4) = 1.5
grams CuCO3 = mols CuCO3 x molar mass CuCO3.
To determine the theoretical yield of malachite when 1.5 moles of Copper Sulphate reacts with an excess of sodium carbonate, we need to balance the chemical equation first. The balanced equation for the reaction between Copper Sulphate (CuSO4) and sodium carbonate (Na2CO3) is as follows:
CuSO4 + Na2CO3 -> CuCO3 + Na2SO4
From the balanced equation, we can see that the stoichiometric ratio between Copper Sulphate and CuCO3 is 1:1. This means that 1 mole of Copper Sulphate will produce 1 mole of CuCO3.
Therefore, the number of moles of CuCO3 produced can be calculated as follows:
Number of moles of CuCO3 = 1.5 moles
To convert the number of moles of CuCO3 to grams, we need to know the molar mass of CuCO3. Copper (Cu) has a molar mass of 63.55 g/mol, while Carbon (C) has a molar mass of 12.01 g/mol, and Oxygen (O) has a molar mass of 16.00 g/mol.
The molar mass of CuCO3 can be calculated as follows:
Molar mass of CuCO3 = (63.55 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
= 187.55 g/mol
Therefore, the mass of CuCO3 produced can be calculated as follows:
Mass of CuCO3 = Number of moles of CuCO3 * Molar mass of CuCO3
= 1.5 moles * 187.55 g/mol
= 281.33 grams
Hence, theoretically, 281.33 grams of malachite should be formed when 1.5 moles of Copper Sulphate reacts with an excess of sodium carbonate.
To determine the theoretical amount of malachite formed when 1.5 moles of Copper Sulphate reacts with excess sodium carbonate, we need to first write the balanced chemical equation for the reaction between Copper Sulphate (CuSO4) and sodium carbonate (Na2CO3).
The balanced equation is as follows:
CuSO4 + Na2CO3 → CuCO3 + Na2SO4
From the balanced equation, we can see that one mole of CuSO4 reacts with one mole of Na2CO3 to produce one mole of CuCO3. Therefore, the stoichiometric ratio between CuSO4 and CuCO3 is 1:1.
Since we know the number of moles of CuSO4 (1.5 moles), we can say that 1.5 moles of CuSO4 would produce 1.5 moles of CuCO3.
Next, we need to convert the moles of CuCO3 to grams using the molar mass of CuCO3.
The molar mass of CuCO3 is calculated by adding the atomic masses of copper (Cu), carbon (C), and oxygen (O):
Cu: 63.55 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (x3, as there are three oxygen atoms in the formula)
Molar mass of CuCO3 = (63.55 g/mol) + (12.01 g/mol) + (16.00 g/mol * 3)
= 123.55 g/mol
Next, we can calculate the theoretical mass of CuCO3 formed by multiplying the number of moles (1.5 moles) by the molar mass (123.55 g/mol):
Mass of CuCO3 = 1.5 moles * 123.55 g/mol
= 185.325 grams
Therefore, theoretically, approximately 185.325 grams of malachite (CuCO3) should be formed when 1.5 moles of Copper Sulphate reacts with an excess of sodium carbonate.