Given that the molar mass of H2O is 18.02 g/mol, how many liters of propane are required at STP to produce 75 g of H2O from this

 <span>(NH4)2CO3 -> 96.09 g/mol

(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate

In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.

Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3

This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.

(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>

Answer:

0.400 moles of Oxygen

Explanation:

By employing the equation PV = nRT, the initial pressure of the flask can be calculated prior to the reaction, which leads to:

P = nRT/V

Where:

n signifies moles (4,000 moles: 2,000 moles of CO and 2,000 moles of H₂O)

R represents the gas constant (0.082 atm·L/mol·K)

T is the temperature (300.0 K)

V denotes volume (0.2000 L)

Substituting values results in P = 492.0 atm

To achieve a pressure reduction of 10.00%, the resulting pressure should be:

492.0 atm - 49.2 atm = 442.8 atm

Calculating with the new pressure under the same conditions gives the moles as:

n = PV/RT

n = 3,600 total moles

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The resulting moles are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

Where X accounts for the moles that react

Consequently, the total moles are:

4,000 moles - X = 3,600 moles

X = 0.400 moles

This indicates that the amount of oxygen needed for the reaction is 0.400 moles of Oxygen

I hope this is useful!

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Initially, we calculate the moles of gas using the ideal gas law:

PV = nRT
n = PV / RT
n = (1.4 * 226.4) / (0.082 *(27 + 273.15))
n = 12.88

Next, we apply the given percentages to estimate the moles of helium:
Moles of helium = 0.655 * 12.88
Moles of helium = 8.44

We then use the formula:
Mass = moles * molar mass

Mass of helium = 8.44 * 4

Mass of helium = 33.76 grams.