use and activity series to Identify two metals that will not generate hydrogen gas when treated with and acid
write balanced molecular equations: aluminum metal with dilute nitric acid
Write balanced molecular equations:Calcium hydroxide solution with acetic acid
How many liters of CO2 form at STP if 5.0g of CaCO3 are treated with excess hydrochloric acid? show all your work
A duplicate post. Responded below.
To identify two metals that will not generate hydrogen gas when treated with an acid, we can refer to the activity series of metals. The metals located below hydrogen in the activity series will not generate hydrogen gas when treated with an acid.
For the first question, aluminum is located above hydrogen in the activity series. Therefore, it will generate hydrogen gas when treated with an acid like dilute nitric acid. The balanced molecular equation for the reaction between aluminum and dilute nitric acid is:
2Al + 6HNO3 -> 2Al(NO3)3 + 3H2
For the second question, calcium hydroxide is a base and not a metal, so we cannot use the activity series in this case. However, calcium hydroxide will react with acetic acid to form water and a salt called calcium acetate. The balanced molecular equation for this reaction is:
Ca(OH)2 + 2CH3COOH -> Ca(CH3COO)2 + 2H2O
For the third question, we can calculate the number of liters of CO2 formed at STP when 5.0g of CaCO3 (calcium carbonate) are treated with excess hydrochloric acid (HCl).
First, we need to write the balanced chemical equation for the reaction between CaCO3 and HCl:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
According to the equation, 1 mole of CaCO3 produces 1 mole of CO2. To find the moles of CaCO3, we use its molar mass:
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 100.09 g/mol
Moles of CaCO3 = 5.0 g / 100.09 g/mol ≈ 0.050 mol
So, 0.050 mol of CaCO3 will produce 0.050 mol of CO2. We can now convert moles of CO2 to liters at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies 22.4 liters.
Liters of CO2 = 0.050 mol x 22.4 L/mol = 1.12 L
Therefore, 5.0 g of CaCO3 treated with excess hydrochloric acid will produce approximately 1.12 liters of CO2 at STP.