Answer:
See attached worksheet
Step-by-step explanation:
As the worksheet explains, the fiorst step is to have a balanced equation of the prioction of hydrogen gas from magnesium and hydrochloric acid.
2Mg + 2HCl ⇒ H2 + 2MgCl
Then convert the mass of manesium into moles Mg (0.045 moles Mg)
Since the balanced equation promises us that we'll get 1 mole of H2 for every 2 moles of Mg, we can predict that the 0.045 moles of Mg will be enough to make (0.045 moles Mg)*(1 mole H2/2 moles Mg) = 0.0225 moles of H2.
At 22.4 l/mole at STP, this means the gas volume of H2 is (0.0225 moles)*(22.4L/mole) = 0.504 L of H2 at STP.
Since the lab is not at STP, the above volume must be adjected for actual lab conditions. Use the combined gas law:
P1V1/T1 = P2V2/T2
where P is pressure, V is volume, and T is temperature, in Kelven. (add 273.15 to C for Kelvin). The subscripts 1 and 2 represent starting (1) and final (2) conditions.
Stand pressure in kPa is 101.3 kPa.
We want V2, so rearrange the equation:
V2 = V1(T2/T1)(P1/P2)
Note that I've arranged the tempaerature and pressure values as ratios. It helps guide the process for cancelling units and understanding the impact of changes in these values.
See the worksheet for more details. V2 is calculated to be 0.542L (542ml) of H2 under the lab conditions.
Percent theorectical yield is the amount obtained divided by the amount expected:
(145ml)/(542ml) = 26.8% theorectical yield.