Question

1. what is the theoretical yield of waffles if you have 5 cups of flour, 9 eggs and 3 tbs of oil?

given: 2 cups flour + 3 eggs + 1 tbs oil - 4 waffles
a) 10
b) 8
c) 12
d) 4
e) not enough information

1. if the theoretical yield of a reaction is 75.0 grams of product and the actual yield is 42.0g. what is

the percent yield?
a) 31.5
b) 178
c) 75.0
d) 56.0
e) none of the above

1. what is the limiting reagent for the following reaction given we have 3.4 moles of ca(no3)2 and 2.4

moles of li3po4?
reaction: 3ca(no3)2 + 2li3po4 - 6lino3 + ca3(po4)2
a) ca3(po4)2
b) ca(no3)2
c) lino3
d) li3po4
e) not enough information

1. how many moles of lithium nitrate are theoretically produced if we start with 3.4 moles of

ca(no3)2 and 2.4 moles of li3po4?
reaction: 3ca(no3)2 + 2li3po4 - 6lino3 + ca3(po4)2
a) 6.8
b) 7.2
c) 1.2
d) 1.1
e) not enough information

1. what is the excess reagent for the reaction below given that you start with 10.0 g of al and

19.0 grams of o2?
reaction: 4al + 3o2 - 2al2o3
a) o2
b) al
c) al2o3
d) both al and o2
e) not enough information

Explanation:

Step1: Determine the number of waffles from each ingredient

• Flour:

Given \(2\) cups flour make \(4\) waffles.
If we have \(5\) cups of flour, the number of waffles \(n_{flour}=\frac{5}{2}\times4 = 10\)

• Eggs:

Given \(3\) eggs make \(4\) waffles.
If we have \(9\) eggs, the number of waffles \(n_{eggs}=\frac{9}{3}\times4=12\)

• Oil:

Given \(1\) tbs oil make \(4\) waffles.
If we have \(3\) tbs oil, the number of waffles \(n_{oil}=\frac{3}{1}\times4 = 12\)

Step2: Identify the limiting ingredient

The limiting ingredient is the one that gives the least number of waffles.
Since \(n_{flour}=10\), \(n_{eggs} = 12\), \(n_{oil}=12\), flour is the limiting ingredient.

Step1: Use the percent - yield formula

The percent - yield formula is \(\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%\)
Given \(\text{Actual Yield}=42.0\space g\) and \(\text{Theoretical Yield}=75.0\space g\)

Step2: Calculate the percent yield

\(\text{Percent Yield}=\frac{42.0}{75.0}\times 100\%=56.0\%\)

Step1: Use the mole ratio from the balanced equation

From the balanced equation \(3Ca(NO_3)_2+2Li_3PO_4
ightarrow6LiNO_3 + Ca_3(PO_4)_2\)
The mole ratio of \(Ca(NO_3)_2\) to \(Li_3PO_4\) is \(\frac{3}{2}\)

Step2: Calculate the amount of \(Li_3PO_4\) needed for \(3.4\) moles of \(Ca(NO_3)_2\)

If \(n_{Ca(NO_3)_2}=3.4\space mol\), the moles of \(Li_3PO_4\) required \(n_{Li_3PO_4}^{req}=\frac{2}{3}\times3.4\space mol\approx2.27\space mol\)

Step3: Compare with the given moles of \(Li_3PO_4\)

We have \(n_{Li_3PO_4}=2.4\space mol\). Since \(2.27\space mol<2.4\space mol\), \(Ca(NO_3)_2\) is the limiting reagent.

Step1: Use the mole ratio from the balanced equation

From \(3Ca(NO_3)_2+2Li_3PO_4
ightarrow6LiNO_3 + Ca_3(PO_4)_2\)
Since \(Ca(NO_3)_2\) is the limiting reagent (\(n_{Ca(NO_3)_2}=3.4\space mol\))
The mole ratio of \(Ca(NO_3)_2\) to \(LiNO_3\) is \(\frac{3}{6}=\frac{1}{2}\)

Step2: Calculate the moles of \(LiNO_3\)

\(n_{LiNO_3}=2\times n_{Ca(NO_3)_2}\)
\(n_{LiNO_3}=2\times3.4\space mol = 6.8\space mol\)

Answer:

A. 10