Chromium (III) oxide, often called chromic oxide, has been used as green paint p

Question

Chromium (III) oxide, often called chromic oxide, has been used as green paint pigment, as a catylistin organic synthesis, as a polishing powder, and to make metallic chromium, One way to make chromium (III) oxide is by reacting sodium dichromate, Na2Cr2O7 with ammonium chloride at 800 to 1000 degrees celcius to form chromium (III) oxide, sodium chloride, nitrogen and water.

a) write a balanced equation for this reaction

b) What is the minimum mass, in megagrams, of ammonium chloride necessary to react completely with 275 Mg of sodium dichromate, Na2Cr2O7?

c) What is the maximum mass, in megagrams, of chromium (III) oxide that can be made from 275 Mg of sodium dichromate, Na2Cr2O7, and excess ammonium chloride?

d) If 147Mg of chromium (III) oxide is formed in the reaction of 275 Mg of sodium dichromate, Na2Cr2O7, with excess ammonium chloride, what is the percent yeild?

Explanation / Answer

Answer – a) We are given reaction between sodium dichromate (Na2Cr2O7) and ammonium chloride form the chromium (III) oxide, sodium chloride, nitrogen and water, so the balanced reaction for this one as follow –

Na2Cr2O7(s) + 2NH4Cl(s) ----> Cr2O3(s) + 2NaCl(s) + N2(g) + 4H2O(g)

b) we are given, mass of mass of Na2Cr2O7(s) = 275 mg

Calculation of moles of Na2Cr2O7(s)- We need the mass in gram, so first we need to convert the mass Mg to g

We know,

1 Mg = 1.0*106 g

So, 275 Mg = ?

= 2.75*108 g

Moles of Na2Cr2O7(s) = 2.75*108 g / 261.97 g.mol-1

                                    = 1.05*106 moles

Moles of ammonium chloride NH4Cl

From the balanced equation-

1 mole of Na2Cr2O7(s) = 2 moles of NH4Cl

So, 1.05*106 moles of Na2Cr2O7(s) = ?

= 2.1*106 moles of NH4Cl

Calculation of ass of NH4Cl –

Mass of NH4Cl = 2.1*106 moles * 53.49 g/mol

                         = 1.12*108 g

Convert the mass g to megagram

We know,

1 g = 1.0*10-6 Mg

So, 1.12*108 g = ?

= 112 Mg

the minimum mass, in megagrams, of ammonium chloride necessary to react completely with 275 Mg of sodium dichromate is 112 Mg.

c) Now we calculated moles of 275 mg of Na2Cr2O7(s)

moles of Cr2O3(s) – from the balanced reaction –

1 mole of Na2Cr2O7(s) = 1 moles of Cr2O3(s)

So, 1.05*106 moles of Na2Cr2O7(s) = ?

= 1.05*106 moles of Cr2O3(s)

Calculation of ass of Cr2O3(s) –

Mass of Cr2O3(s) = 1.05*106 moles * 151.99 g/mol

                             = 1.60*108 g

Convert the mass g to megagram

We know,

1 g = 1.0*10-6 Mg

So, 1.60*108 g = ?

= 160 Mg

the maximum mass, in megagrams, of chromium (III) oxide that can be made from 275 Mg of sodium dichromate is 160 Mg.

d) we are given, actual mass of Cr2O3(s) = 147 mg

we calculated the theoretical yield = 160 Mg

we know,

Percent yield = actual yield / theoretical yield * 100 %

                      = 147 Mg / 160 Mg

                      = 91.8 %.

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