For the reaction shown, calculate how many grams of each product form when the given amount of each reactant completely reacts to form products. Assume there is more than enough of the other reactant.
2 HCl(aq) + Na2CO3(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)
(a) 10.8 g HCl
(b) 10.8 g Na2CO3
Here, we are going to calculate the mass of product formed when the given amount of each reactant completely reacts.
The given reaction is:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
Given, mass of HCl = 10.8 g
Molar mass of HCl = 36.46 g/mol
Therefore, number of moles of HCl = given mass / molar mass
= = 0.296 mol
From the above reaction, it is seen that 2 moles of HCl produces 2 moles of NaCl, 1 mole of H2O and 1 mole of CO2.
Therefore, 0.296 mol of HCl will produce 0.296 mol of NaCl, (½ x 0.296 = 0.148) mole of H2O and 0.148 mole of CO2.
Now, 1 mol NaCl = molar mass of NaCl
Therefore, 0.296 mol of NaCl = 0.296 x molar mass of NaCl
= 0.296 x 58.44 g/mol
= 17.3 g
Again, 1 mol H2O = molar mass of H2O
Therefore, 0.148 mol of H2O = 0.148 x molar mass of H2O
= 0.148 x 18.01 g/mol
= 2.7 g
Again, 1 mol CO2 = molar mass of CO2
Therefore, 0.148 mol of CO2 = 0.148 x molar mass of CO2
= 0.148 x 44.01 g/mol
= 6.5 g
Thus 10.8 g HCl produces 17.3 g NaCl, 2.7 g H2O and 6.5 g CO2.