If 235.7 g of NaClO3 is heated, what is the total mass of the products?
The balanced equation for the decomposition of sodium chlorate is:
2 NaClO3 → 2 NaCl + 3 O2
From the balanced equation, we can see that 2 moles of NaClO3 decompose to form 2 moles of NaCl and 3 moles of O2.
To find the mass of the products, we need to calculate the number of moles of NaClO3. We can use the molar mass of NaClO3 to convert grams to moles.
Molar mass of NaClO3 = 22.99 g/mol (Na) + 35.45 g/mol (Cl) + 3 * 16.00 g/mol (O) = 106.44 g/mol
Number of moles of NaClO3 = mass of NaClO3 / molar mass of NaClO3 = 235.7 g / 106.44 g/mol = 2.215 mol
Since the ratio of NaClO3 to NaCl is 2:2 (from the balanced equation), the number of moles of NaCl will be equal to the number of moles of NaClO3.
Number of moles of NaCl = 2.215 mol
Since the ratio of NaClO3 to O2 is 2:3 (from the balanced equation), the number of moles of O2 will be equal to 3/2 times the number of moles of NaClO3.
Number of moles of O2 = (3/2) * 2.215 mol = 3.3225 mol
To find the mass of the products, we multiply the number of moles of each product by their respective molar masses.
Mass of NaCl = number of moles of NaCl * molar mass of NaCl = 2.215 mol * (22.99 g/mol + 35.45 g/mol) = 160.85 g
Mass of O2 = number of moles of O2 * molar mass of O2 = 3.3225 mol * 32.00 g/mol = 106.6 g
Total mass of the products = mass of NaCl + mass of O2 = 160.85 g + 106.6 g = 267.45 g
Therefore, the total mass of the products is 267.45 g.