Dr.Bob pls check my work. I'm not sure if im right
C(s) + 2 H2(g) → CH4(g)
A. Write the mole ratio for C to H2
2mol H2/ 1mol C
B. If a total of 3 moles C and 4 moles H2 are available, which reactant is limiting?
3 moles/1 = 3 moles C, 4 mols H2/2 = 2 mols H2. Since H2 has smaller value, H2 is limiting.
C. Using the amounts in part b, what is the maximum number of moles of CH4 that can be formed?
2 moles H = 1 mole CH4
2:1 ratio
Therefore: 4 moles H2 * 2/1 = 8 moles of CH4
D. If 12 grams of carbon and 12 grams of hydrogen are available, what is the maximum mass of methane that can be produced?
2:1 ratio
12 grams H2 * 2/1 = 24 grams
24 grams methane produced
A is fine. B is correct but I don't think the explanation is the best. Both C and D are wrong.
C(s) + 2 H2(g) → CH4(g)
B. If a total of 3 moles C and 4 moles H2 are available, which reactant is limiting?
3 moles/1 = 3 moles C, 4 mols H2/2 = 2 mols H2. Since H2 has smaller value, H2 is limiting.
I would prefer that you had done it this way. You can get confused your way sometimes. Proof of that lies in your answer for C and it isn't clear whether to use mole ratio of 1/2 or 2/1.
3 mols C produces 3 mols CH4 since 3 mols C x (1 mol CH4/1 mol C) = 3 x 1/1 = 3 mols CH4. 4 mols H2 produces 4 x (1 mol CH4/2 mols H2) = 4 x 1/2 = 2 mols CH4. H2 is the LR because it will produce the smaller amount. </
C. Using the amounts in part b, what is the maximum number of moles of CH4 that can be formed?
2 moles H = 1 mole CH4
2:1 ratio
Therefore: 4 moles H2 * 2/1 = 8 moles of CH4
You goofed on this because you took a shortcut and didn't use the units. The correct way is 4 mols H2 x (1 mol CH4/2 mols H2) = 4 x 1/2 = 2 mol CH4. The problem said to use the data from part B and you didn't. You calculated in part B that you would get 2 mols CH4 from 4 mols H2. Actually you didn't need to calculate anything because you already knew the answer from part B; i. e., 2 mols CH4.
D. Part D is a repeat of part B EXCEPT part B gave the information in moles. Part D, however, gives the information in grams. You must convert grams to mols for each material BEFORE you calculate it as I've shown for part B.