CH4(g) + 4S(g)> CS2(g) + 2H2S(g)

The given problem doesn't have anything to do with Cl2, so I am assuming that you have 75.5 g of S instead.
First off, convert the given quantity of CH4 and S to moles.
Carbon has a molar mass of 12.01 g/mol and Hydrogen has a molar mass of 1.01 g/mol, so the molar mass of CH4 is just the sum of the molar mass of a Carbon atom and 4 Hydrogen atoms, which is 16.05 g/mol. Sulfur just has a molar mass of 64.07 g/mol.
Dividing the quantities of the compounds given by their respective molar mass gives:
35.8/16.05 = 2.23 mol CH4 and 75.5/64.07 = 1.18 mol S.
If we let Sulfur go into access, 2.23 mol of CH4 will produce 2(2.23) (<== from the balanced equation) = 4.46 mol of H2S. If Methane goes into access, then (2/4)(2.23) = 1.12 mol of H2S is produced. This shows that Methane is the limiting reagent and 1.12 mol of H2S is produced.
Now, just convert this quantity to grams and you'll get your answer.
I hope this helps!
First off, convert the given quantity of CH4 and S to moles.
Carbon has a molar mass of 12.01 g/mol and Hydrogen has a molar mass of 1.01 g/mol, so the molar mass of CH4 is just the sum of the molar mass of a Carbon atom and 4 Hydrogen atoms, which is 16.05 g/mol. Sulfur just has a molar mass of 64.07 g/mol.
Dividing the quantities of the compounds given by their respective molar mass gives:
35.8/16.05 = 2.23 mol CH4 and 75.5/64.07 = 1.18 mol S.
If we let Sulfur go into access, 2.23 mol of CH4 will produce 2(2.23) (<== from the balanced equation) = 4.46 mol of H2S. If Methane goes into access, then (2/4)(2.23) = 1.12 mol of H2S is produced. This shows that Methane is the limiting reagent and 1.12 mol of H2S is produced.
Now, just convert this quantity to grams and you'll get your answer.
I hope this helps!

I think you ment S, since that's the only other reactant
Divide the masses by their molar masses to get moles...
35.8gCH4 x (1molCH4/16.0426gCH4)=2.23mol CH4
75.5g S x (1mol Cl2/ 32.065gCl2)=2.35mol S
CH4/S=1/4 this is the molar ratio in the equation
2.23/(4*2.23)=2.23/8.92=CH4/S, the thing is, we don't have8.92mol of S, this is called a limited reactant, so let's look at the other conversion factor...
S/CH4=4/1 this is the molar ration seen in the equation
2.35/(2.35/4)=2.35/.5875=S/CH4, so the max of CH4 is 2.23, and we only need .5875mol of CH4, we can finally compare this to the H2S, which has two moles so...
CH4/H2S=1/2=.5875/(.5875*2)=.5875/1.17…
1.175 mol H2S x (34.0808gH2S/1mol H2S)=40.04g H2S
Divide the masses by their molar masses to get moles...
35.8gCH4 x (1molCH4/16.0426gCH4)=2.23mol CH4
75.5g S x (1mol Cl2/ 32.065gCl2)=2.35mol S
CH4/S=1/4 this is the molar ratio in the equation
2.23/(4*2.23)=2.23/8.92=CH4/S, the thing is, we don't have8.92mol of S, this is called a limited reactant, so let's look at the other conversion factor...
S/CH4=4/1 this is the molar ration seen in the equation
2.35/(2.35/4)=2.35/.5875=S/CH4, so the max of CH4 is 2.23, and we only need .5875mol of CH4, we can finally compare this to the H2S, which has two moles so...
CH4/H2S=1/2=.5875/(.5875*2)=.5875/1.17…
1.175 mol H2S x (34.0808gH2S/1mol H2S)=40.04g H2S

Can we assume you meant S and Not Cl2?
Step 1: Balanced Equation. CH4(g) + 4S(g)> CS2(g) + 2H2S(g)
Step 1: Balanced Equation. CH4(g) + 4S(g)> CS2(g) + 2H2S(g)
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keywords: grams,produced,of,are,How,mixture,75.5,If,35.8,many,and,reacts,CH,Cl,If a mixture of 35.8 grams of CH4 and 75.5 g of Cl2 reacts, How many grams of H2S are produced