1. Write the balanced equation for the reaction conducted in this lab, including appropriate phase symbols. Mg(s) + 2HCl(aq) –> H2(g) + MgCl2(aq) 2. Determine the partial pressure of the hydrogen gas collected in the gas collection tube. The partial pressure of the hydrogen gas is 1.07 atm 3. Calculate the moles of hydrogen gas collected. pv=mrt ; n= .0013mol of hydrogen gas 4. If magnesium was the limiting reactant in this lab, calculate the theoretical yield of the gaseous product. Show all steps of your calculation. 0.03184 g Mg(1mol Mg/ 24.3050 g Mg)= 0.0013mols Mg
0.0013mols Mg(1mol H2/ 1mol Mg)= .0013
1. Determine the percent yield of this reaction, showing all steps of your calculation.
2. (actual yield/ theoretical yield)x 100%
(0.0013/ 0.0013)x 100= 100%
1. Would the following errors increase, decrease, or have no effect on the calculated moles of gas collected in the experiment? Explain your answers in complete sentences. a. The measured mass of the magnesium was smaller than the true mass. If the measured amount of magnesium is less than the true amount of magnesium then the measured number of moles would be lower than the measured number of hydrogen would be smaller than the true number of moles of hydrogen. It would decrease the calculated moles of the gas. b. The actual temperature of the hydrogen gas is lower than room temperature. If the actual temperature of hydrogen gas is lower than room temperature then the actual amount of hydrogen gas would be higher than the gas at room temperature. So therefore it would increase the number of moles produce.
2. Explain in terms of particle collisions and Dalton’s law why it can be assumed that the total pressure inside the gas collection tube is equal to the atmospheric pressure outside of the tube. If the number of particle collisions and the pressure inside the gas collection tube was high it would force the water in the gas collection tube to be lower than the surrounding water. If the number of particle collisions and the pressure was low the surrounding water would push the water higher into the gas collection tube. When the water is even, inside and outside of the tube, the number of particle collisions and the pressure are equal.
3. If an undetected air bubble was trapped inside the gas collection tube, how would this affect your calculated percent yield? Explain your answer.
4. If there was an undetected air bubble trapped in the gas collection tube then the pressure of hydrogen would be lower then the actual value therefore the percent yield would be lower than the actual percent yield.