All categories

3 years ago

What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?

Chemistry

2 answers:

yawa3891 [41]3 years ago

5 0

The balanced chemical reaction will be:

2H2O = 2H2 + O2

We are given the amount of water used in the decomposition reaction. This will be our starting point.

17.0 g H2O (1 mol H2O/ 18.02 g H2O) (1 mol O2/2 mol H2O) ( 32.00 g O2/1mol O2) = 15.09 g O2

Percent yield = actual yield / theoretical yield x 100

Percent yield =10.2 g / 15.09 g x 100

Percent yield = 67.58%

sineoko [7]3 years ago

3 0

Answer: D

Explanation: 67.6

You might be interested in

If a microgram is one millionth of a gram (0.000 00 1 g), how many micrograms are there in one gram?

den301095 [7]

Answer:

There are One million micrograms in a gram

Explanation:

A microgram is one Millionth of a gram therefore it take one million of them to equal one gram

7 0

3 years ago

Aspirin can be made in the laboratory by reacting acetic anhydride (c4h6o3) with salicylic acid (c7h6o3) to form aspirin (c9h8o4

Temka [501]

V(C₄H₆O₃) = 5.00 mL.
d(C₄H₆O₃) = 1.08 g/mL.
m(C₄H₆O₃) = V(C₄H₆O₃) · d(C₄H₆O₃).
m(C₄H₆O₃) = 5.00 mL · 1.08 g/mL.
m(C₄H₆O₃) = 5.4 g.
n(C₄H₆O₃) = m(C₄H₆O₃) ÷ M(C₄H₆O₃).
n(C₄H₆O₃) = 5.4 g ÷ 102 g/mol.
n(C₄H₆O₃) = 0.0529 mol.
n(C₇H₆O₃) = 2.08 g ÷ 138.1 g/mol.
n(C₇H₆O₃) = 0.015 mol; limiting reactant.
From chemical reaction: n(C₄H₆O₃) : n(C₉H₈O₄) = 1 : 1.
n(C₉H₈O₄) = 0.015 mol.
m(C₉H₈O₄) = 0.015 mol · 180.16 g/mol.
m(C₉H₈O₄) = 2.71 g; theoretical yield.
percent yield od aspirine = 2.57 g ÷ 2.71 g · 100% = 94.83%.

7 0

3 years ago

Which subshell holds the valence electrons of barium?.

olga2289 [7]

Answer:

Explanation:

Barium is in group 2 of the s block and is in period 6.

7 0

2 years ago

Why is the solvation energy negative? a) bonds are being created b) bonds are breaking

IrinaVladis [17]

Answer:

your answer will be b) bonds are breaking

3 0

3 years ago

. What is a carbonyl group? Draw the carbonyl groups that are characteristic of aldehydes, ketones, carboxylic acids, and esters

Elan Coil [88]

Answer :

Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as $C=O$

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.

The general formula of carboxylic is, $C_nH_{2n+1}COOH$ . According to the IUPAC naming, the carboxylic are named as alkanoic acids.

Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.

The general representation of aldehyde is, $R-CHO$ . According to the IUPAC naming, the aldehyde are named as alkanals.

Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.

The general representation of ketone is, $R-CO-R^'$ . According to the IUPAC naming, the ketone are named as alkanone.

Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.

The general representation of ester is, $R-COO-R^'$ . According to the IUPAC naming, the ester are named as alkyl alkanoate.

6 0

3 years ago