Question

1. For the reaction C + 2H2 → CH4, how many grams of carbon are required to produce 19.5 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

2. For the reaction C + 2H2 → CH4, how many moles of carbon are needed to make 119.4 grams 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


3. 2 NH3 + 3 CuO --> 3 Cu + N2 + 3 H2O

In the above equation how many moles of N2 can be made when 127.4 grams of CuO 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

Answer 1

1) 234,0 g 2) 7,5 3) 0,5

Step-by-step explanation:

1) Since the stoichiometric coefficients of carbon and methane are equal, the moles of carbon needed are the same as the moles of methane produced. Therefore the mass of carbon needed can be calculated as follows

`m = 19.5 * 12 = 234.0`

2)This exercise is comparable to the former one. Since the stoichiometric coefficients are the same, the moles of methane procured and the moles of the carbon needed are the same.

The molar mass of methane is 12 + 1 × 4 = 16

`n_(C) = (119.4 g)/(16) =7.5`

3) In this case, the stoichiometric coefficients are not equal. In order to produce 1 mole of nitrogen, 3 moles of CuO (copper (II) oxide) are needed. Therefore, the number of moles of CuO consumed must be divided by 3 in order to get the moles of nitrogen produced.

Molar mass of CuO = 63,5 + 16 = 79.5

`n_CuO = (127,4 g)/(79,5) = 1,6 mol \\ n_(N2) = (1,6)/(3) = 0,5 mol`