Asked by Lena
                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)?
            
        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)?
Answers
                    Answered by
            DrBob222
            
    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
    
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
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