Imagine that it is lunch time and you need to make a total of four turkey sandwi
ID: 991580 • Letter: I
Question
Imagine that it is lunch time and you need to make a total of four turkey sandwiches for you and your friends. The "formula" for making a turkey sandwich calls for two slices of bread, two slices of turkey, and one slice of cheese. To make the four sandwiches, you will need eight slices of bread, eight slices of turkey, and four slices of cheese. When you go to the refrigerator and gather all of the ingredients, you discover that you only have seven slices of bread instead of the eight that you need. You have plenty of the rest of the ingredients. Since you don't have enough bread to make all four sandwiches, the bread is considered to be the limiting reactant. A limiting reactant will determine how much product, in this case the sandwich, you can make. It is the reactant, or ingredient, that is present in the smallest amount when you take into consideration the ratio of reactants. The rest of the ingredients are considered to be excess reactants, meaning you have more than is necessary. Based on the amount of ingredients you have and the "formula", you will be able to make three and one-half sandwiches. You will use all of the bread and will have leftover turkey and cheese.
1.How many moles of PCl5 can be produced from 26.0 g of P4 (and excess Cl2)? Express your answer to three significant figures and include the appropriate units.
2.How many moles of PCl5 can be produced from 57.0 g of Cl2 (and excess P4)? Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
1) How many moles of PCl5 can be produced from 26.0 g of P4 (and excess Cl2)?
P4 + 10 Cl2 --------------> 4 PCl5
1 mol = 124 g 4 mol
26 g ?
? = ( 26 g/124 g ) x 4 mol of PCl5
= 0.84 mol of PCl5
Therefore,
0.84 moles of PCl5 can be produced from 26.0 g of P4 (and excess Cl2).
2) How many moles of PCl5 can be produced from 57.0 g of Cl2 (and excess P4)?
P4 + 10 Cl2 --------------> 4 PCl5
10 mol
= 10 x 71 g = 710 g 4 mol
57 g ?
? = ( 57 g/710 g ) x 4 mol of PCl5
= 0.32 mol of PCl5
Therefore,
0.32 moles of PCl5 can be produced from 57.0 g of Cl2 (and excess P4).