How many milliliters of 0.710 M HCl are required to react with 47.80 grams of CaCO3 ?
Given the balanced reaction equation:
6 FeCl2(aq) + K2Cr2O7(aq) + 14 HCl(aq)
6 FeCl3(aq) + 2 CrCl3(aq) + 2 KCl(aq) + 7 H2O(l)
How many moles of FeCl2(aq) are required to produce 2.811 mol of FeCl3(aq)?
For #1, write the equation.
2HCl + CaCO3 ==> CaCl2 + H2O + CO2
moles CaCO3 = grams/molar mass
Convert moles CaCO3 to moles HCl using the coefficients in the balanced equation (which is why you need the equation).
Now convert moles HCl to volume using M = moles/L and convert L to mL.
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)
Stomach acid is essentially 0.10 M HCl. An active ingredient found in a number of popular antacids is CaCO3. Calculate the number of grams of CaCO3 needed to exactly react with 250 mL of stomach acid. CaCO3(s)+ 2HCl(aq)= CO2(g)+