N2(g) + 3H2(g) ------------> 2NH3(g) no of moles of N2 = W/G.M.Wt = 2.8/28 = 0.1
ID: 700016 • Letter: N
Question
N2(g) + 3H2(g) ------------> 2NH3(g)
no of moles of N2 = W/G.M.Wt
= 2.8/28 = 0.1mole
no of moles of H2 = W/G.M.Wt
= 1/2 = 0.5 moles
from balanced equation
3 moles of H2 react with 1 mole of N2
0.5 moles of H2 react with = 1*0.5/3 = 0.166 moles of N2 is required
N2 is limiting reactant
2.
N2(g) + 3H2(g) ------------> 2NH3(g)
1 mole of N2 react with H2 to gives 2 moles of NH3
0.1 mole of N2 react with H2 to gives = 2*0.1/1 = 0.2 moles of NH3
mass of NH3 = no of moles * gram molar mass
= 0.2*17 = 3.4g
b. 3.4g >>>>>>answer
c.
N2(g) + 3H2(g) ------------> 2NH3(g)
1 mole of N2 react with 3 moles of H2
0.1 mole of N2 react with = 3*0.1/1 = 0.3 moles of H2
H2 is excess reactant
The no of moles of H2 remains = 0.5-0.3 = 0.2 moles of H2
mass of H2 remains = no of moles * gram molar mass
= 0.2*2 = 0.4 g of excees of H2
d.0.4g of H2
Explanation / Answer
27.(8pts total) We start with 2.80 g of N2 and 1.00 g of Hz to form NH. Start by writing and balancing the reaction.) (1) (2 pts) Which is the limiting reactant? (a) H2 (b) N2 (2) (3 pts) How many grams of NH, would be formed? (b) 3.40g (c) 9.10g (d) 12.3g (e) none of these (a) 1.70g (3) (3 pts) What is left over and how many grams? (a) 0.200 g ofN2 (b) 0.400 g of N2 (c) 0.200 g of H2 (d) 0.400 g of H2 (e) none of these
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