When 1.53 grams of silver metal are dropped in 500. mL of 2.0 M acetic acid, how many liters of gas are evolved at STP?
See your post above on HCN.
Let me show you my work.
2Ag+2C2H4O2=2AgC2H3O2+H2
I found M of Ag 0.0142 to give me M of H 7.09x10-3
when I plug it back into the equation Pv=nRT I got .0158 L. I put it in as my answer and it tells me I forgot to check something. What am i doing wrong?
To solve this problem, we need to determine the amount of a gas evolved in the reaction between silver metal and acetic acid, and then convert it to liters at STP (Standard Temperature and Pressure).
The balanced chemical equation for this reaction is:
2Ag + 2CH3COOH -> Ag2(CH3COO)2 + H2
From the balanced equation, we can see that 2 moles of silver (Ag) react to produce 1 mole of hydrogen gas (H2).
First, let's calculate the number of moles of silver (Ag) present in 1.53 grams using its molar mass of 107.87 g/mol:
moles of Ag = mass of Ag / molar mass of Ag
= 1.53 g / 107.87 g/mol
≈ 0.0142 mol
Since 2 moles of Ag react to produce 1 mole of H2, we have:
moles of H2 = 0.0142 mol / 2
= 0.0071 mol
Now, to convert the moles of H2 to volume at STP, we use the ideal gas law:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume of gas
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (STP = 273.15 K)
Rearranging the equation, we have:
V = nRT / P
Plugging in the values, we get:
V = (0.0071 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)
≈ 0.153 L
Therefore, when 1.53 grams of silver metal are dropped in 500 mL of 2.0 M acetic acid, approximately 0.153 liters of gas are evolved at STP.