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) ?

Question

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) ?

Damon · Answer

.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

DrBob222 · Answer

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.