You have two typos in your equation. The correct equation, balanced, is
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
mols CaCO3 = grams/molar mass
mols CO2 = mols CaCO3 from the coefficients in the balanced equation.
Now g CO2 = mols CO2 x molar mass CO2.
Calcium carbonate CaCO3 reacts with stomach acid (HCl, hydrochloric acid) according to the following equation: CaCO3(s)+2HCl(aq)->CaCl2(aq)+H2O(l)+CO2(g) Tums, an antacid, contains CaCO3. If Tums is added to 35.0 mL of 0.300 M
Calcium carbonate reacts w/stomach acid according to the following chemical equation. CaCO3+2HCl(aq)-> CaCl2(aq)+H2O(l)+CO2(g) A. Balance the eqaution B. Tums is one commercially sold antacid that contains CACO3. If Tums is added
How many grams of hydrochloric acid are required to react completely with 4.30g of zinc? How many molecules of gas will be produced? 1. Write the balanced chemical equation. Zn + 2HCl ==> ZnCl2 + H2 2. Convert grams Zn to mols.
CaCO3(s)+2HCl(aq)-->CaCl2(aq)+H2O(l)+CO2 (g) How many grams of calcium chloride will be produced when 31g of calcium carbonate are combined with 11 g of HCl? I have 0.3097 moles of CaCO3 and 0.30169 moles of HCl...I get stuck at
A 0.450-gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is: CaCO3(s)+2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g) The excess HCl(aq) is titrated by 9.75 mL of 0.125 M NaOH(aq).
How many milliliters of 0.418 M HCl are needed to react with 52.7 g of CaCO3? 2HCl (aq) + CaCO3 (s) >>> CaCl2 (aq) + CO2 (g) + H20 (l) I got .00252 mL but its wrong. This is what I did: 52.7 g x (1 mol CaCO3/100.09 CaCO3) x (2 mol