10. What do you have to do to the coefficients of equation I below to get to equation II? i. 2 SnO2 + 4 H → 2 Sn + 4 H 0
ii. Sno, + 2 H, → Sn + 2 H2O
Can you divide equation II by another factor and still have it be correct? Why or why not

Magnesium is the limiting reactant in this experiment. calculate the theoretical yield of mgo for each trial. trial 1: trial 2: data mass of empty crucible with lid trial 1: 26.688 trial 2: 26.681 mass of mg metal, crucible, and lid trial 1: 26.994 trial: 2 26.985 mass of mgo, crucible, and lid trial 1: 27.188 trial 2: 27.180