In the following equations determine which reactant is the limiting and which reaactant is in excess.
1. KCL + HNO3--->KNO3 + HCL
22.0g 18.3g
2. 2Al(OH)3 + 3H2SO4---> Al2(SO4)3 + 6H2O where there is 25 g of Al(OH)3 and 25g H2SO4.
You can do these one of two ways. One is the short way and the other a much longer way (but easier to explain). Here is the short way.
1. Balance the equation.'
2. Convert grams to moles of each reagent.
3. Run a ratio of the two and see which is the limiting reagent. Here is how to do the first one but I'll just be approximate. You will need to go through and redo it by not rounding the molar masses etc.
The first thing, however, is that the first reaction you have does not take place. However, I assume this is an exercise on limiting reagents and we will assume the reaction occurs.
KCl + HNO3 ==> HCl + KNO3
1. The equation is balanced.
2. Convert grams to moles.
moles KCl = 22.0 g/molar mass KCl = 22.0/about 75 = 0.293 moles KCl.
moles HNO3 = 18.3/about 63 = 0.290
Since the mole ratio of the reactants is 1:1 (meaning that 1 mole KCl reacts with 1 mole HNO3), then the smaller number of moles will be the limiting reagent. In this case, HNO3 is the smaller which means that 0.290 moles of HNO3 will react with 0.290 moles KCl and some of the KCl will remain un-reacted. HNO3 is the limiting reagent. The second problem is a little more complicated since they aren't 1:1. We can help you through it if you get stuck but post your work and explain what you don't understand it you need additional assistance.