Under certain circumstances, carbon dioxide, CO2(g), can be made to react with h

Question

Under certain circumstances, carbon dioxide, CO2(g), can be made to react with hydrogen gas,H2(g), to produce methane, CH4(g), and water vapor, H2O(g):

CO2(g)+4H2(g)?CH4(g)+2H2O(g)

1. How many moles of methane are produced when 48.1moles of carbon dioxide gas react with excess hydrogen gas?

Express your answer with the appropriate units. For example, write the unit moles as mol.

2.How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 76.6moles of water vapor?

Express your answer with the appropriate units. For example, write the unit moles as mol.

_____________________________________________________________________________________

The reaction of hydrogen with oxygen produces water.
2H2(g)+O2(g)?2H2O(g)

1.How many moles of O2 are required to react with 4.0moles H2?

2.If you have 9.0moles O2, how many moles of H2 are needed for the reaction?

3.How many moles of H2O form when 5.5moles O2 reacts?

_______________________________________________________________

Carbon disulfide and carbon monoxide are produced when carbon is heated with sulfur dioxide.
5C(s)+2SO2(g)?CS2(l)+4CO(g)

1. How many moles of C are needed to react with 0.550mole SO2?

2.How many moles of CO are produced when 1.9moles C reacts?

3.How many moles of SO2 are required to produce 0.75mole CS2?

4.How many moles of CS2 are produced when 2.1moles C reacts?

_______________________________________________________________________

Sodium reacts with oxygen to produce sodium oxide.
4Na(s)+O2(g)?2Na2O(s)

1. How many grams of Na2O are produced when 42.4g of Na reacts?

2.If you have 21.0g of Na, how many grams of O2 are required for the reaction?

3.How many grams of O2 are needed in a reaction that produces 73.0g of Na2O?

Explanation / Answer

A. .1) 48.1 mol CO2 * (1 mol CH4 / 1 mol CO2) = 48.1 mol CH4

2) 88.1 mol H2O * (4 mol H2 / 2 mol H2O) = 176.2 mol H2

B. Balanced equation:
2H2 + O2 ? 2H2O
2mol H2 reacts with 1 mol O2 to produce 2 mol H2O
1) 8mol H2 will react with 8/2 = 4 mol O2

2) 18 mol of H2 will react with 9 mol O2

3) 1mol O2 will produce 2 mol H2O
3.50 mol O2 will produce 3.5*2 = 7.0 mol H2O

C your S is a 5

0.550 *5/2

2.2 * 4/5

0.35 *2

2.5 * 1/5

D.

22g of Na is 22/23 moles of Na
1/4 that number is the number of moles of O2 needed
Each mole of O2 weighs 32g

mass of O2 needed is thus 1/4 * 22/23 * 32 = 7.65g

the stoichiometry also tells us that 1/2 * 22/23 moles of the oxide are produced.
Each mole of Na2O weighs 62g
So mass of the oxide produced is 1/2 * 22/23 * 62 = 29.65g

(the calculation confirms what could have been calculated much easier , simply by adding the mass of the reactants together!)

Rather than 29.65g of the oxide we want 74.0g
This is a 74.0/29.65 multiple.
So the mass of O2 must also go up by the same multiple
7.65 * 74.0/29.65 = 19.09

19.09g of O2 are needed.

.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

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