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.
For #2, here is a link to solve stoichiometry problems. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the number of moles of FeCl2 required to produce a certain number of moles of FeCl3, we need to look at the balanced reaction equation.
From the equation, we can see that the stoichiometric ratio between FeCl2 and FeCl3 is 6:6, which means that for every 6 moles of FeCl2, we will produce 6 moles of FeCl3.
So, to solve this problem, we can set up a proportion to find the moles of FeCl2:
(6 moles FeCl2 / 6 moles FeCl3) = (x moles FeCl2 / 2.811 moles FeCl3)
Cross-multiplying and solving for x, we get:
6 moles FeCl2 * 2.811 moles FeCl3 = 6 moles FeCl3 * x moles FeCl2
16.866 moles FeCl3 = 6x
Dividing both sides by 6, we find:
x = 16.866 moles FeCl3 / 6
x = 2.811 moles FeCl2
So, 2.811 moles of FeCl2 are required to produce 2.811 moles of FeCl3.