All categories

3 years ago

If the theoretical yield of RX is 56.0 g , what is the percent yield ?

Chemistry

1 answer:

maks197457 [2]3 years ago

4 0

the complete question in attached figure

Let
x------------------- > actual yield
y------------------- > theoretical yield
z------------------- > percent yield
we have that
z=x/y

we know
x=47 g
y=56 g

therefore

z=47/56=0.839 ---------------- > 83.9%

the answer is the option C 83.9%

You might be interested in

0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2Al(OH)3(s) +

Veseljchak [2.6K]

Answer:

The answer to your question is 1.1 moles of water

Explanation:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

0.45 mol 0.55 mol ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

3 moles of H₂SO₄ ---------------- 6 moles of water

0.55 moles of H₂SO₄ ----------- x

x = (0.55 x 6) / 3

x = 3.3 / 3

x = 1.1 moles of water

5 0

3 years ago

Help PLZ!!! Which statement accurately represents the arrangement of electrons in Bohr’s atomic model? -Electrons move around th

Novosadov [1.4K]

Answer:

electrons move around the nucleus in fixed orbits of equal amounts of energy

Explanation:

Quantum Theory

7 0

3 years ago

Could someone help me with this?

Varvara68 [4.7K]

Solving part-1 only

KMnO_4

Transition metal is Manganese (Mn)

Actually it's the oxidation number of Mn

Let's find how?

$\\ \tt\Rrightarrow x+1+4(-2)=0$

$\\ \tt\Rrightarrow x+1-8=0$

$\\ \tt\Rrightarrow x-7=0$

$\\ \tt\Rrightarrow x=+7$

x is the oxidation number

Purple as per the color of potassium permanganate

$\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}$