1. Part B 0.30 mole of C2H6 Express your answer using two significant figures. P

Question

1. Part B

0.30 mole of C2H6 Express your answer using two significant figures.

Part D

What mass of PCl5 will be produced from the given masses of both reactants? Express your answer to three significant figures and include the appropriate units. Part A=0.872 mole ans., Part C= 0.316 Mole ans.

2.Part D

P4+5O22P2O5

What is the percent yield if the actual yield from this reaction is 194 g ? Express your answer to three significant figures and include the appropriate units. Part A=2.60 mol, Part B=2.20 mol, Part C=2.20 mol

3. In the body, the amino acid glycine, C2H5NO2, reacts according to the following equation:

2C2H5NO2(aq) +3O2(g) 3CO2(g)+3H2O(l)+CH4N2O(aq)

Glycine                                                                             Urea

Part A

How many grams of O2 are needed to react with 15.5 g of glycine?

Express your answer with the appropriate units.

mO2=

mol

Part B

How many grams of urea are produced from 15.5 g of glycine?

Express your answer with the appropriate units.

murea=

M

Part C

How many grams of CO2 are produced from 15.5 g of glycine?

Express your answer with the appropriate units.

mCO2=

4. Calculate the number of moles in 3.47 g of each of the following:

Part A

He

Express your answer with the appropriate units.

n=

Part B

SnO2

Express your answer with the appropriate units.

n=

Part C

Cr(OH)3

Express your answer with the appropriate units.

n=

Part D

Ca3N2

Express your answer with the appropriate units.

n=

Balance each of the following by determining coefficients, and identify the type of reaction.

Part A

Enter your answers numerically separated by commas.

Part B

Identify the type of the reaction from Part A:

combination

decomposition

single replacement

double replacement

combustion

Part C

Enter your answers numerically separated by commas.

Part D

Identify the type of the reaction from Part C:

combination

decomposition

single replacement

double replacement

combustion

mO2=

mol

Explanation / Answer

2.PART D:

P4 + 5O2 --------> 2P2O5

3×30.97g 5×32g 2× 141.94g

As per the stoichiometry 283.88 gm of P2O5 should be the output for 100 yield . So , if 194g P2O5 is the out put them the percentage yield is 194/283.88×100 = 68.34

3. 2C2H5NO2 +. 3O2 -------- > 3CO2 + 3H2OO + CH4N2O

2×75.06g 3×32g 3× 44g. 3×18g. 60.06g

150.12g 96g 132g 54g 60.06g

PARTA: 96g of O2 needed to react with 150.12g of glycine .Therefore to react with 15.5g of glycine O2 Need is 96/150.12 × 15.5 = 9.29g of O2 required

Part B: From 150.12 g of glycine 60.06g of urea produced so from 15.5g of glycine ,60.06/150.12×15.5 = 6.06g of urea produced.

part c:From 150.12g of glycine 132 g of CO2 produced ,So from 15.5 g of glycine 132/150.12×15.15g = 13.63g of CO2 produced.

4.PartA : 0.867 moles of He

part b : 0.0230 moles of SnO2

Part c: 0.0336 moles of Cr(OH)3

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