#1. Once made by heating wood in the absence of air, methanol (CH3OH) is now mad

Question

#1. Once made by heating wood in the absence of air, methanol (CH3OH) is now made by reacting carbon monoxide and hydrogen at high pressure: CO(g)+2H2(g)CH3OH(l)

A. If 25.0 g of CO is reacted with 6.00 g of H2, which is the limiting reagent?

B. How many grams of CH3OH can be made from 14.78 g of CO and 5.321 of H2?

#2. For the reaction 4Al(s)+3O2(g)2Al2O3(s), What is the mass of Al2O3 that is produced when 24.25 g of Al and 24.25 g of O2 react

#3. For the reaction 3NO2(g)+H2O(l)2HNO3(aq)+NO(g),

What is the mass of HNO3 that is produced when 28.05 g of NO2 and 28.05 g of H2O react?

#4. For each of the following precipitation reactions, calculate how many grams of the first reactant are necessary to completely react with 56.1 g of the second reactant.

A. 2KI(aq)+Pb(NO3)2(aq)PbI2(s)+2KNO3(aq)

B. 3 Na2CO3(aq)+2FeCl3(aq)Fe2(CO3)3(s)+6NaCl(aq)

C. K2SO4(aq)+Sr(NO3)2(aq)SrSO4(s)+2KNO3(aq)

Explanation / Answer

1. Once made by heating wood in the absence of air, methanol (CH3OH) is now made by reacting carbon monoxide and hydrogen at high pressure: CO(g)+2H2(g)CH3OH(l)

A. If 25.0 g of CO is reacted with 6.00 g of H2, which is the limiting reagent?

Solution :- Lets calculate the mass of the H2 needed to react with the 25.0 g CO

(25.0 g CO * 1 mol CO / 28 g)*(2 mol H2/ 1 mol CO)*(2.015 g/1 mol H2) =3.59 g H2

Mass of the H2 needed for the reaction is more therefore the H2 is the excess reactant hence the CO is the limiting reactant.

B. How many grams of CH3OH can be made from 14.78 g of CO and 5.321 of H2?

Solution :- Using the mole ratio of the both reactants lets calculate the mass of the product formed from each

(14.78 g CO*1 mol/28 g)*(1 mol CH3OH/1 mol CO)*(32.04 g/1 mol CH3OH) = 16.9 g CH3OH

(5.321 g H2*1mol/2.015g)*(1 mol CH3OH/2mol H2)*(32.04 g/1 mol CH3OH)= 42.3 g CH3OH

Mass of the CH3OH that can be produced = 16.9 g because CO is the limiting reactant

#2. For the reaction 4Al(s)+3O2(g)2Al2O3(s), What is the mass of Al2O3 that is produced when 24.25 g of Al and 24.25 g of O2 react

Solution :- using the mole ratio of both reactant calculate the mass of the Al2O3

(24.25 g Al*1 mol/26.982 g)*(2molAl2O3 / 4 mol Al)*(101.96 g / 1 molAl2O3) =45.8 g Al2O3

(24.25g O2*1mol/32 g)*(2mol Al2O3/3molO2)*(101.96 g/ 1 mol Al2O3)= 51.5 g Al2O3

Mass of the Al2O3 that can be produced = 45.8 g because Al is the limiting reactant.

#3.  For the reaction 3NO2(g)+H2O(l)2HNO3(aq)+NO(g),

What is the mass of HNO3 that is produced when 28.05 g of NO2 and 28.05 g of H2O react?

Solution :- Lets calculate the mass of the HNO3 that can be produced from each reactant

(28.05 g NO2*1 mol / 46.0 g)*(2molHNO3/3mol NO2)*(63.01g /1mol HNO3) = 25.6 g HNO3

(28.05 g H2O*1 mol/18.015g)*(2mol HNO3/1mol H2O)*(63.01 g/1 mol HNO3) = 196 g HNO3

Mass of the HNO3 that can be produced is 25.6 g because NO2 is the limiting reactant

#4. For each of the following precipitation reactions, calculate how many grams of the first reactant are necessary to completely react with 56.1 g of the second reactant.

A. 2KI(aq)+Pb(NO3)2(aq)PbI2(s)+2KNO3(aq)

Solution :-

(56.1 g Pb(NO3)2*1mol/331.2g)*(2 mol KI/1mol Pb(NO3)2)*(166 g/1mol KI) = 56.2 g KI

So mass of KI needed = 56.2 g

B. 3 Na2CO3(aq)+2FeCl3(aq)Fe2(CO3)3(s)+6NaCl(aq)

Solution :-

(56.1 g FeCl3 *1 mol / 162.2g)*(1 mol Na2CO3/2 molFeCl3)*(106 g / 1 mol Na2CO3)= 18.3 g

So the mass of Na2CO3 needed = 18.3 g

C. K2SO4(aq)+Sr(NO3)2(aq)SrSO4(s)+2KNO3(aq)

Solution :-

(56.1 g Sr(NO3)2 * 1 mol / 211.63 g)*(1 mol K2SO4/ 1 mol Sr(NO3)2)*(174.26 g/1mol K2SO4) = 46.2 g

So the mass of the K2SO4 needed = 46.2 g

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.