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
I ahve the identical questions...Is your book Modern Chemistry?
How about you showing all your work and we will help you with it? Hint for the first one. Look at the acitivity series of metals and pick two that are below hydrogen.
what two? metals that will not generate hydrogen gas when treated with an acid
To identify two metals that will not generate hydrogen gas when treated with an acid, we can consult an activity series. The activity series ranks metals in order of their reactivity with acids, from most reactive to least reactive.
In a typical activity series, the most reactive metals are at the top, and the least reactive metals are at the bottom. If a metal appears below hydrogen in the series, it means that metal will not generate hydrogen gas when treated with an acid.
Two common metals that will not generate hydrogen gas when treated with an acid are silver and gold.
Now, let's write the balanced molecular equations for the given reactions:
1. Aluminum metal with dilute nitric acid:
The balanced molecular equation is:
2 Al (s) + 6 HNO3 (aq) → 2 Al(NO3)3 (aq) + 3 H2O (l)
2. Calcium hydroxide solution with acetic acid:
The balanced molecular equation is:
Ca(OH)2 (aq) + 2 CH3COOH (aq) → Ca(CH3COO)2 (aq) + 2 H2O (l)
Finally, let's calculate the amount of CO2 formed when 5.0g of CaCO3 are treated with excess hydrochloric acid at STP.
The balanced chemical equation for this reaction is:
CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
First, we need to calculate the number of moles of CaCO3:
Number of moles = mass / molar mass
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol
Number of moles of CaCO3 = 5.0 g / 100.09 g/mol ≈ 0.050 mol
According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CO2.
Therefore, the number of moles of CO2 formed is also 0.050 mol.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.
Hence, the volume of CO2 formed will be:
Volume of CO2 = 0.050 mol × 22.4 L/mol ≈ 1.12 L
Therefore, 5.0g of CaCO3 will produce approximately 1.12 liters of CO2 at STP.