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 21)

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:

19)

Step1: Calculate waffles from flour

Given \(2\) cups flour make \(4\) waffles. For \(5\) cups flour: \(\frac{4}{2}\times5 = 10\) waffles.

Step2: Calculate waffles from eggs

Given \(3\) eggs make \(4\) waffles. For \(9\) eggs: \(\frac{4}{3}\times9=12\) waffles.

Step3: Calculate waffles from oil

Given \(1\) tbs oil make \(4\) waffles. For \(3\) tbs oil: \(\frac{4}{1}\times3 = 12\) waffles.
The limiting reactant is flour (gives least waffles).

Step1: Use percent - yield formula

Percent yield \(=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\). Given actual yield \(=42.0\) g, theoretical yield \(=75.0\) g.
Percent yield \(=\frac{42.0}{75.0}\times 100=56.0\)

Step1: Use mole - ratio from reaction

From \(3Ca(NO_3)_2+2Li_3PO_4
ightarrow6LiNO_3 + Ca_3(PO_4)_2\), mole ratio \(Ca(NO_3)_2:Li_3PO_4 = 3:2\)
For \(3.4\) moles \(Ca(NO_3)_2\), moles of \(Li_3PO_4\) needed \(=\frac{2}{3}\times3.4\approx2.27\) moles. We have \(2.4\) moles \(Li_3PO_4\).
For \(2.4\) moles \(Li_3PO_4\), moles of \(Ca(NO_3)_2\) needed \(=\frac{3}{2}\times2.4 = 3.6\) moles. We have \(3.4\) moles \(Ca(NO_3)_2\).
\(Ca(NO_3)_2\) is limiting.

Answer:

A. 10

20)