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Methane and hydrogen sulfide form when 36g H2reacts with carbon disulfide. Calculate the percent yield if the actual yield of CH4is 69.8g.
4H2(g) + CS2(g) CH4(g) + 2H2S(g)
a.
97%
c.
85%
b.
94%
d.
72%
**Weegy:** First we have to find the theoretical yield by doing a regular stoichiometric problem:
Given g H2 x 1 mol H2/molar mass H2 x molar ratio(mol CH4/mol H2) x molar mass CH4/1 mol CH4 = __ g CH4
Molar mass H2:
2 H x 1.008 = 2.016 g/mol
Molar mass [ CH4:
1 C x 12.01 = 12.01
4 H x 1.008 = 4.032
Add them up, we get 16.042 g/mol
The mole ratio can be obtained from the coefficients of the balanced equation: 1 mol CH4 : 4 mol H2
36 g H2 x (1 mol H2/2.016 g H2) x (1 mol CH4/4 mol H2) x (16.042 g CH4/1 mol CH4) = 71.6 g CH4
Now we find the percent yield by doing: mass actual yield/mass of theoretical yield x 100%
69.8/71.6 x 100% = 97.486%
Therefore , when Methane and hydrogen sulfide form when 36g H2reacts with carbon disulfide. The percent yield if the actual yield of CH4is 69.8g is a. 97% . ] (More)

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