Given the reaction: Fe2SiO4 (s) + 2 H2CO3 (aq) ----> 2FeCO3 (s) + H4SiO4 (aq). How many grams of Fe2SiO4 are required to completely react with 55.0 mL of .500 M H2CO3 (aq)?
To solve this problem, we need to use stoichiometry and the concept of limiting reactants. The balanced equation tells us that the molar ratio between Fe2SiO4 and H2CO3 is 1:2.
First, let's find the number of moles of H2CO3 in 55.0 mL of 0.500 M H2CO3 (aq). We can use the formula:
moles of solute = concentration (M) x volume (L)
Converting mL to L:
volume = 55.0 mL / 1000 mL/L = 0.055 L
Now, we can find the moles of H2CO3:
moles of H2CO3 = 0.500 M x 0.055 L = 0.0275 mol
Since the molar ratio between Fe2SiO4 and H2CO3 is 1:2, we need twice as many moles of Fe2SiO4 to react completely with the given moles of H2CO3. Therefore, we need 0.0275 mol x 2 = 0.055 mol of Fe2SiO4.
Finally, to find the mass of Fe2SiO4 needed, we need to multiply the number of moles by its molar mass. The molar mass of Fe2SiO4 can be calculated using the atomic masses of the elements:
Fe: 55.85 g/mol
Si: 28.09 g/mol
O: 16.00 g/mol
Molar mass of Fe2SiO4 = (2 x 55.85 g/mol) + 28.09 g/mol + (4 x 16.00 g/mol) = 231.93 g/mol
Now, we can calculate the mass of Fe2SiO4:
mass = moles x molar mass = 0.055 mol x 231.93 g/mol = 12.76 g
Therefore, approximately 12.76 grams of Fe2SiO4 are required to completely react with 55.0 mL of 0.500 M H2CO3 (aq).