## Chemistry 10th Edition

There are 882. g of $NaHCO_3$
1. Find the number of moles of $NaHCO_3$: Each $NaHCO_3$ has 1 carbon. So: $10.5mol(C) \times \frac{1mol(NaHCO_3)}{1mol(C)} = 10.5 mol(NaHCO_3)$ 2. Determine the molar mass of this compound $(NaHCO_3)$: 22.99* 1 + 1.008* 1 + 12.01* 1 + 16.00* 3 = 84.01g/mol 3. Calculate the mass ($NaHCO_3$) $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $84.01 * 10.5 = mass(g)$ $882. = mass(g)$