According to the following reaction, how many grams of dinitrogen monoxide will

Question

According to the following reaction, how many grams of dinitrogen monoxide will be formed upon the complete reaction of 32.8 grams of ammonium nitrate? B. According to the following reaction, how many grams of chlorine gas are required for the complete reaction of 32.1 grams of iron? iron (s) + chlorine (g) iron(III) chloride (s) C. An iron nail rusts when exposed to oxygen. For the following reaction, 3.09 grams of iron are mixed with excess oxygen gas . The reaction yields 3.69 grams of iron(III) oxide . iron ( s ) + oxygen ( g ) iron(III) oxide ( s ) What is the theoretical yield of iron(III) oxide ? grams What is the percent yield for this reaction ? D.For the following reaction, 3.86 grams of hydrochloric acid are mixed with excess iron. The reaction yields 6.19 grams of iron(II) chloride. iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g) What is the theoretical yield of iron(II) chloride ? grams What is the percent yield of iron(II) chloride ? % E. You need to make an aqueous solution of 0.123 M calcium nitrate for an experiment in lab, using a 250 mL volumetric flask. How much solid calcium nitrate should you add? How many milliliters of an aqueous solution of 0.148 M nickel(II) iodide is needed to obtain 18.6 grams of the salt? F. You wish to make a 0.220 M hydroiodic acid solution from a stock solution of 3.00 M hydroiodic acid. How much concentrated acid must you add to obtain a total volume of 150 mL of the dilute solution? In the laboratory you dilute 5.29 mL of a concentrated 12.0 M hydrobromic acid solution to a total volume of 50.0 mL. What is the concentration of the dilute solution? G.How many grams of PbBr2 will precipitate when excess CrBr3 solution is added to 75.0 mL of 0.557 M Pb(NO3)2 solution? 3Pb(NO3)2(aq) + 2CrBr3(aq) 3PbBr2(s) + 2Cr(NO3)3(aq) How many mL of 0.580 M HI are needed to dissolve 9.53 g of BaCO3? 2HI(aq) + BaCO3(s) BaI2(aq) + H2O(l) + CO2(g) I. An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of oxygen gas are necessary to form 0.169 moles iron(III) oxide? iron (s) + oxygen (g) iron(III) oxide (s) According to the following reaction, how many moles of hydrochloric acid are necessary to form 0.332 moles iron(II) chloride? iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g)

Explanation / Answer

A) According to the following reaction, how many grams of dinitrogen monoxide will be formed upon the complete reaction of 32.8 grams of ammonium nitrate?

the reaction is not given, but I hope the reaction is:

NH4NO3 -------> N2O + 2H2O

Mass of NH4NO3 used = 32.8 g

Moles of NH4NO3 = mass/molar mass of NH4NO3 = 32.8 g/(80 g/mol) = 0.41 mol

1 mol of NH4NO3 produces 1 mol of N2O.

So, 0.41 mol of NH4NO3 produces 0.41 mol of N2O.

Mass of N2O produced = Number of moles*molar mass of N2O = 0.41*44 = 18.04 g

Mass of N2O produced = 18.04 g

B) According to the following reaction, how many grams of chlorine gas are required for the complete reaction of 32.1 grams of iron? iron (s) + chlorine (g) iron(III) chloride (s)

iron (s) + chlorine (g) iron(III) chloride (s)

2Fe + 3Cl2 ---------> 2FeCl3

Mass of Fe used = 32.1 g

Moles of Fe = mass/molar mass of Fe = 32.1 g/(55.8 g/mol) = 0.575 mol

2 mol of Fe produces 2 mol of FeCl3.

So, 0.575 mol of Fe produces 0.575 mol of FeCl3.

Mass of FeCl3 produced = Number of moles*molar mass of FeCl3 = 0.575*162.2 = 93.27 g

Mass of FeCl3 produced = 93.27 g

C) An iron nail rusts when exposed to oxygen. For the following reaction, 3.09 grams of iron are mixed with excess oxygen gas . The reaction yields 3.69 grams of iron(III) oxide . iron ( s ) + oxygen ( g ) iron(III) oxide ( s ) What is the theoretical yield of iron(III) oxide ? grams What is the percent yield for this reaction ?

iron ( s ) + oxygen ( g ) iron(III) oxide ( s )

4Fe + 3O2 --------> 2Fe2O3

Mass of Fe used = 3.09 g

Moles of Fe = mass/molar mass of Fe = 3.09 g/(55.8 g/mol) = 0.055 mol

4 mol of Fe produces 2 mol of Fe2O3.

So, 0.055 mol of Fe produces Fe2O3 = (2/4)*0.055 = 0.027 mol

Mass of Fe2O3 produced = Number of moles*molar mass of Fe2O3 = 0.027*160 = 4.43 g

Theoretical yield of Fe2O3 produced = 4.43 g

Experimental yield = 3.69 g

% yield = (experimental/theoretical)*100 = (3.69/4.43)*100 = 83.3 %

D) For the following reaction, 3.86 grams of hydrochloric acid are mixed with excess iron. The reaction yields 6.19 grams of iron(II) chloride. iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g) What is the theoretical yield of iron(II) chloride ? grams What is the percent yield of iron(II) chloride ? %

iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g)

Fe + 2HCl --------> FeCl2 + H2

Mass of HCl used = 3.86 g

Moles of HCl = mass/molar mass of HCl = 3.68 g/(36.5 g/mol) = 0.101 mol

2 mol of HCl produces 1 mol of FeCl2.

So, 0.101 mol of HCl produces FeCl2 = (1/2)*0.101 = 0.05 mol

Mass of FeCl2 produced = Number of moles*molar mass of FeCl2 = 0.05*126.8 = 6.34 g

Theoretical yield of FeCl2 produced = 6.34 g

Experimental yield = 6.19 g

% yield = (experimental/theoretical)*100 = (6.19/6.34)*100 = 97.6 %

E) E. You need to make an aqueous solution of 0.123 M calcium nitrate for an experiment in lab, using a 250 mL volumetric flask. How much solid calcium nitrate should you add?

Molarity = 0.123 M

Volume = 250 ml

Molar mass of Ca(NO3)2 = 164 g

Molarity = (mass*1000)/(molar mass*volume)

0.123 = (mass*1000)/(164*250)

mass = 5.043 g

Mass of Ca(NO3)2 required = 5.043 g

How many milliliters of an aqueous solution of 0.148 M nickel(II) iodide is needed to obtain 18.6 grams of the salt?

Molarity = 0.148 M

Mass = 18.6 g

Molar mass of NiI2 = 312.5 g

Molarity = (mass*1000)/(molar mass*volume)

0.148 = (18.6*1000)/(164*volume)

volume = 766.3 ml

Aqueous solution required = 766.3 ml

F) You wish to make a 0.220 M hydroiodic acid solution from a stock solution of 3.00 M hydroiodic acid. How much concentrated acid must you add to obtain a total volume of 150 mL of the dilute solution?

M1 = 0.220 M

V1 = 150 ml

M2 = 3.00 M

V2 = ?

M1V1 = M2 V2

V2 = M1V1/M2 = 0.220*150/3.00 = 11 ml

volume of concentrated acid required = 11 ml

In the laboratory you dilute 5.29 mL of a concentrated 12.0 M hydrobromic acid solution to a total volume of 50.0 mL. What is the concentration of the dilute solution?

M1 = 12.0 M

V1 = 5.29 ml

M2 = ?

V2 = 50.0 ml

M1V1 = M2 V2

M2 = M1V1/V2 = 12.0*5.29/50.0 = 1.27 M

concentration of the dilute solution = 1.27 M

G.How many grams of PbBr2 will precipitate when excess CrBr3 solution is added to 75.0 mL of 0.557 M Pb(NO3)2 solution?

3Pb(NO3)2(aq) + 2CrBr3(aq) 3PbBr2(s) + 2Cr(NO3)3(aq)

Molarity = 0.557 M

Volume = 75.0 ml

Molar mass of Pb(NO3)2 = 331.2 g/mol

Molarity = (mass*1000)/(molar mass*volume)

0.557 = (mass*1000)/(331.2*75)

mass = 13.84 g

Mass of Pb(NO3)2 required = 13.84 g

Moles of Pb(NO3)2 = mass/molar mass of Pb(NO3)2 = 13.84 g/(331.2 g/mol) = 0.042 mol

3 mol of Pb(NO3)2 produces 3 mol of PbBr2.

So, 0.042 mol of Pb(NO3)2 produces 0.042 mol of PbBr2.

Mass of PbBr2 produced = Number of moles*molar mass of PbBr2 = 0.042*367 = 15.33 g

Mass of PbBr2 produced = 15.33 g

H) How many mL of 0.580 M HI are needed to dissolve 9.53 g of BaCO3? 2HI(aq) + BaCO3(s) BaI2(aq) + H2O(l) + CO2(g)

Mass of BaCO3 = 9.53 g

moles of BaCO3 = mass/molar mass of BaCO3 = 9.53 g/(197.3 g/mol) = 0.05 mol

1 mol of BaCO3 require 2 mol of HI.

0.05 mol of BaCO3 requires 0.10 mol of HI

Molarity = number of moles/volume(L)

volume(L) = number of moles/molarity = 0.10/0.580 = 0.172 L = 172 ml

Volume of HI = 172 ml

I) An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of oxygen gas are necessary to form 0.169 moles iron(III) oxide? iron (s) + oxygen (g) iron(III) oxide (s)

2Fe + 3O2 --------> 2Fe2O3

2 mol of Fe2O3 is produced from 3 mol of O2.

So, 0.169 mol of Fe2O3 is produced from O2 = (3/2)*0.169 = 0.254 mol

moles of oxygen gas are necessary = 0.254 mol

J) According to the following reaction, how many moles of hydrochloric acid are necessary to form 0.332 moles iron(II) chloride? iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g)

Fe + 2HCl --------> FeCl2 + H2

For 1 mol of FeCl3, 2 mol of HCl is required.

For 0.332 mol of FeCl3, HCl required is = 2*0.332 = 0.664 mol

HCl required = 0.664 mol

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