if 3.85 L of carbon monoxide was collected over water at 25 degrees celsius and 689 mmHg how many grams of formic acid were consumed?

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4.69g HCOOH consumed?

To find the number of grams of formic acid consumed, we need to know the stoichiometry of the reaction between carbon monoxide (CO) and formic acid (HCOOH). The balanced chemical equation for this reaction is as follows:

2 CO + H2O -> HCOOH + CO2

From the equation, we see that for every 2 moles of carbon monoxide, we get 1 mole of formic acid.

Given that we have collected 3.85 L of carbon monoxide gas, we first need to find the number of moles of CO. We can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (689 mmHg)
V = volume (3.85 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (25 + 273.15 K)

Substituting the values into the equation, we can solve for n:

(689 mmHg) * (3.85 L) = n * (0.0821 L·atm/(mol·K)) * (25 + 273.15 K)

Simplifying the equation:

(689 mmHg * 3.85 L) = n * (0.0821 L·atm/(mol·K)) * (298.15 K)

2649.65 mmHg·L = n * 24.502735 L·mmHg/(mol·K)

Dividing both sides by (24.502735 L·mmHg/(mol·K)):

n = (2649.65 mmHg·L) / (24.502735 L·mmHg/(mol·K))

n ≈ 108.0877 mol

Since the stoichiometry for the reaction is 2 moles of CO to 1 mole of HCOOH, we have:

108.0877 mol CO × (1 mol HCOOH / 2 mol CO) = 54.04385 mol HCOOH

Now, to find the number of grams of formic acid consumed, we need to multiply the number of moles by the molar mass of formic acid (HCOOH).

The molar mass of formic acid (HCOOH) is approximately 46.03 g/mol.

54.04385 mol HCOOH * 46.03 g/mol = 2485.814 g

Therefore, approximately 2485.814 grams of formic acid were consumed.

To determine the grams of formic acid consumed, we need to use the ideal gas law and the stoichiometry of the reaction between carbon monoxide (CO) and formic acid (HCOOH).

The balanced equation for the reaction is:
CO + H2O -> HCOOH

First, let's convert the volume of carbon monoxide (CO) collected to moles. We can use the ideal gas law equation:

PV = nRT

Where:
P = pressure in atm (689 mmHg / 760 mmHg/atm = 0.904 atm)
V = volume in liters (3.85 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (25 °C + 273.15 = 298.15 K)

Plugging in the values:
(0.904 atm) * (3.85 L) = n * (0.0821 L·atm/mol·K) * (298.15 K)

Solving for n, the number of moles of CO:
n = (0.904 atm * 3.85 L) / (0.0821 L·atm/mol·K * 298.15 K) ≈ 0.147 moles

Since the stoichiometry of the reaction is 1:1 between CO and formic acid, we know that the moles of formic acid consumed are also 0.147 moles.

Now, to calculate the grams of formic acid consumed, we need to multiply the number of moles by the molar mass of formic acid, which is 46.03 g/mol.

Grams of formic acid = 0.147 moles * 46.03 g/mol ≈ 6.76 grams

Therefore, approximately 6.76 grams of formic acid were consumed.