if 2.7 moles od sodium carbonate reacts with 5 moles of copper(II) sulfate, how many grams of sodium sulfate will be produced?
To find out how many grams of sodium sulfate will be produced, we need to use the stoichiometry of the balanced chemical equation for the reaction. Let's start by writing the balanced equation for the reaction between sodium carbonate and copper(II) sulfate:
Na2CO3 + CuSO4 -> Na2SO4 + CuCO3
From the balanced equation, we can see that it takes one mole of sodium carbonate (Na2CO3) to react with one mole of copper(II) sulfate (CuSO4) to produce one mole of sodium sulfate (Na2SO4).
Given that 2.7 moles of sodium carbonate is reacting, we can conclude that 2.7 moles of sodium sulfate will be produced.
Now, we need to convert the moles of sodium sulfate to grams using its molar mass. The molar mass of sodium sulfate (Na2SO4) can be calculated as follows:
2 (atomic mass of sodium, Na) + 1 (atomic mass of sulfur, S) + 4 (atomic mass of oxygen, O)
= 2(22.99 g/mol) + 1(32.07 g/mol) + 4(16.00 g/mol)
= 46.00 g/mol + 32.07 g/mol + 64.00 g/mol
= 142.07 g/mol
Now, we can calculate the grams of sodium sulfate produced:
Grams = Moles x Molar Mass
Grams = 2.7 moles x 142.07 g/mol
Grams = 383.199 g
Therefore, approximately 383.2 grams of sodium sulfate will be produced.