Question

Given 67.0g of UO2 and excess HF, how many L of H2O gas can you make at 766mm of Hg at 245 Celsius?

UO2 + 4HF --- UF4 + 2H2O

Answer 1

Step 1: Explanation

The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation of the behavior of many gases under many conditions.

We know the ideal gas equation, PV = nRT

where, R=universal gas constant ( it may vary in unit like 0.08206 L-atm mol^-1 K^-1 )

T=Temperature ( Unit = Kelvin )

P=Pressure ( Unit= atm )

V=Volume ( Unit=Litre )

n=moles ( Unit = mol )

Step 1 :Write the balanced chemical equation

UO₂ + 4 HF -------> UF₄ + 2 H₂O

Step 2 : Calculate the moles of H₂O

we know,

moles=mass given / molar mass

mass = 67 g

molar mass of UO₂ = ( 238 g+ (2 ×16 g) =  270 g /mol

Moles (n) of UO₂   = 67 g / 270 g/mol = 0.24815 mol

UO₂ + 4 HF -------> UF₄ + 2 H₂O

From the equation we can see

1 moles of UO₂ produced 2 moles of H₂O

so 0.24815  mol of UO₂ will produced  = ( 2 mol / 1 mol )  × 0.24815 mol = 0.4963 mol of H₂O

Step 3: calculate the volume of H₂O

moles(n) = 0.4963 mol

Pressure(P) = 761 mmHg × (1 atm / 760 mmHg ) = 1.0079 atm [ Note: 1 atm = 760 mmHg]

Temperature(T)= 245°C = ( 245 +273) K = 518 K

By using ideal gas equation

PV = nRT

=> V = nRT / P = ( 0.4963  mol  × 0.08206 L.atm /mol.K × 518 K ) / 1.0079 atm = 20.9 L

hence, the volume of H₂O produced = 20.9 L