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)?
1.29g
1. You have the balanced equation.
2. mols H2CO3 = M x L = ?
3. Using the coefficients in the balanced equation, convert mols H2CO3 to mols Fe2SiO4.
4. Now convert mols Fe2SiO4 to grams. g = mols x molar mass = ?
To determine the grams of Fe2SiO4 required to completely react with the given volume and concentration of H2CO3, we'll follow these steps:
Step 1: Write the balanced chemical equation.
The balanced chemical equation is:
Fe2SiO4 (s) + 2 H2CO3 (aq) ----> 2 FeCO3 (s) + H4SiO4 (aq)
Step 2: Calculate the moles of H2CO3.
Using the given volume (55.0 mL) and concentration (0.500 M) of H2CO3, we can convert the volume into moles:
Moles of H2CO3 = Volume (L) x Concentration (M)
First, convert the volume to liters by dividing by 1000:
55.0 mL = 55.0 / 1000 = 0.055 L
Now, calculate the moles:
Moles of H2CO3 = 0.055 L x 0.500 M = 0.0275 moles
Step 3: Use the stoichiometry of the balanced equation.
From the balanced equation, we see that 1 mole of Fe2SiO4 reacts with 2 moles of H2CO3. So the moles of Fe2SiO4 required will be half of the moles of H2CO3 since the ratio is 1:2.
Moles of Fe2SiO4 = 0.0275 moles / 2 = 0.01375 moles
Step 4: Calculate the molar mass of Fe2SiO4.
Fe2SiO4 has a molar mass of:
Fe: 55.85 g/mol x 2 = 111.70 g/mol
Si: 28.09 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total molar mass = 111.70 g/mol + 28.09 g/mol + 64.00 g/mol = 203.79 g/mol
Step 5: Calculate the mass of Fe2SiO4.
Mass of Fe2SiO4 = Moles of Fe2SiO4 x Molar mass of Fe2SiO4
Mass of Fe2SiO4 = 0.01375 moles x 203.79 g/mol = 2.80 grams
Therefore, approximately 2.80 grams of Fe2SiO4 are required to completely react with 55.0 mL of 0.500 M H2CO3 (aq).