I have two reactions.
1) Mg (s) + 2HCl (aq) --> MgCl2 (aq) + H2 (g)
2) MgO (s) + 2HCl (aq) --> MgCl2 (aq) + H2O (l)
Assuming I have 1.0 g of each solid and the concentration of the HCl is 1.0M, how would I find the amount of HCl (in mL) that is needed to complete the reaction?
Any help would be greatly appreciated.
1) Mg (s) + 2HCl (aq) --> MgCl2 (aq) + H2 (g)
2) MgO (s) + 2HCl (aq) --> MgCl2 (aq) + H2O (l)
Assuming I have 1.0 g of each solid and the concentration of the HCl is 1.0M, how would I find the amount of HCl (in mL) that is needed to complete the reaction?
Any help would be greatly appreciated.
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Mg + 2 HCl → MgCl2 + H2
(1.0 g Mg) / (24.3051 g/mol) x (2/1) / (1.0 mol/L) = 0.082 L = 82 mL HCl
MgO + 2 HCl → MgCl2 + H2O
(1.0 g MgO) / (40.3045 g MgO/mol) x (2/1) / (1.0 mol/L) = 0.050 L = 50 mL HCl
(1.0 g Mg) / (24.3051 g/mol) x (2/1) / (1.0 mol/L) = 0.082 L = 82 mL HCl
MgO + 2 HCl → MgCl2 + H2O
(1.0 g MgO) / (40.3045 g MgO/mol) x (2/1) / (1.0 mol/L) = 0.050 L = 50 mL HCl