If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reaction if you captured the gas in a 2.50 L vessel at 30 C

35.1% is percent yield

Explanation:

Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

Moles Ni (Molar mass: 58.69g/mol):

1.02g * (1mol / 58.69g) = 0.01738moles Ni

Moles HBr:

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

Where P is pressure (0.0910atm)

V is volume (2.50L)

R is gas constant (0.082atmL/molK)

T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)

And n are moles

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =