Given the reaction: Fe2SiO4 (s) + 2 H2CO3 (aq) ---> 2 FeCO3 (s) + H4SiO4 (aq) how many grams of Fe2SiO4 (s) are required to completely react with 55.0 mL of .500 M H2CO3 (aq) ?

.055*.5 = .0275 mols of H2CO3

You need half as many mols of Fe2SiO4
.5*.0275 = .01375 mols
now add up gram molecular mass of Fe+Fe+Si+16+16+16+16 to get grams/mol
and multiply by .01375 to get grams

I worked this problem for you this morning at 10:59 A.M. and here you've posted it again about seven hours later. Don't mess up the boards with superfluous questions.

To find the number of grams of Fe2SiO4 required to completely react with the given amount of H2CO3, we need to follow these steps:

Step 1: Write and balance the chemical equation:
Fe2SiO4 (s) + 2 H2CO3 (aq) ---> 2 FeCO3 (s) + H4SiO4 (aq)

Step 2: Calculate the number of moles of H2CO3:
To do this, we can use the formula:
moles = concentration (M) × volume (L)

Given:
Concentration (H2CO3) = 0.500 M (Molarity)
Volume (H2CO3) = 55.0 mL = 0.0550 L (convert mL to L)

So, moles of H2CO3 = 0.500 M × 0.0550 L = 0.0275 moles

Step 3: Use the stoichiometry of the reaction to determine the number of moles of Fe2SiO4:
According to the balanced chemical equation, the stoichiometric ratio between Fe2SiO4 and H2CO3 is 1:2. That means 1 mole of Fe2SiO4 reacts with 2 moles of H2CO3.

Therefore, moles of Fe2SiO4 = 0.0275 moles ÷ 2 = 0.0138 moles

Step 4: Convert moles of Fe2SiO4 to grams of Fe2SiO4:
To convert moles to grams, we need the molar mass of Fe2SiO4.

Molar mass of Fe2SiO4 = (2 × atomic mass of Fe) + atomic mass of Si + 4 × (atomic mass of O)

Using the periodic table, we find:
Atomic mass of Fe = 55.845 g/mol
Atomic mass of Si = 28.0855 g/mol
Atomic mass of O = 15.999 g/mol

Molar mass of Fe2SiO4 = (2 × 55.845 g/mol) + 28.0855 g/mol + (4 × 15.999 g/mol) = 231.395 g/mol

Now we can calculate the mass of Fe2SiO4:
mass (Fe2SiO4) = moles (Fe2SiO4) × molar mass (Fe2SiO4)
mass (Fe2SiO4) = 0.0138 moles × 231.395 g/mol = 3.19 grams (rounded to two decimal places)

Therefore, approximately 3.19 grams of Fe2SiO4 are required to completely react with 55.0 mL of 0.500 M H2CO3.