Question

3 Cu + 8HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O

In the above equation, how many grams of water can be made when 15.4 moles of HNO3 are consumed?



Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element Molar Mass
Hydrogen 1
Nitrogen 14
Copper 63.5
Oxygen 16

Question 2
For the reaction C + 2H2 → CH4, how many grams of carbon are required to produce 5.7 moles of methane, CH4 ?

Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element Molar Mass
Hydrogen 1
Carbon 12

Answer 1

Question 1:

The balanced equation is:

3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O

From the equation, we see that for every 8 moles of HNO3, 4 moles of H2O are produced.

So, for 15.4 moles of HNO3, we can calculate the moles of water produced as follows:

Moles of H2O = (15.4 mol HNO3 / 8 mol HNO3) x 4 mol H2O

Moles of H2O = 7.7 mol H2O

Now, we can use the molar mass of water (18 g/mol) to calculate the mass of water produced:

Mass of H2O = 7.7 mol H2O x 18 g/mol

Mass of H2O = 138.6 g

Rounding to the nearest tenth, the answer is 138.6 g of water.

Therefore, when 15.4 moles of HNO3 are consumed, 138.6 g of water can be made.

Question 2:

The balanced equation is:

C + 2 H2 → CH4

From the equation, we see that for every 1 mole of C, 1 mole of CH4 is produced.

So, for 5.7 moles of CH4, we need 5.7 moles of C.

Now, we can use the molar mass of carbon (12 g/mol) to calculate the mass of carbon required:

Mass of C = 5.7 mol C x 12 g/mol

Mass of C = 68.4 g

Rounding to the nearest tenth, the answer is 68.4 g of carbon.

Therefore, 68.4 g of carbon are required to produce 5.7 moles of methane