How many grams SO2 can be formed from 20.0 g of S and 160g O2?

Question

Please show work so I can understand it thank u.

Answer 1

The equation is S + O2 ===> SO2. That says that 1 g-atom of S reacts with 1 mol of O2 to give 1 mol SO2. The first step is to find the limiting reagent--the one that runs out first.

20g S x (1 g-atomS/32g S) = 0.675 g-atomS

160g O2 x (1 molO2/32g O2) = 5 molO2

So the S runs out first; that is the limiting reagent.

So 0.675 g-atomS x (1 mol SO2/1 g-atomS) x (64gSO2/1 molSO2) = 5/8 x 64 = 40g SO2

Answer 2

Number of moles of S in 20g = 20/32 (AW of S)
= 0.625 m
Number of moles of O2 in 160g = 160/32(AW of O2)
= 5.0 m
O2 will be in excess so the most that can be produced is 0.625 m of SO2
Atomic Weight of SO2 = 64 (32 +16+16)
0.625m x 64 = 40g

So your answer is 40g

Answer 3

S(s) + O2(g) --> SO2(g)

The S and O2 remain in their proportions, therefore the mass of SO2 produced = 20 + 160 = 180g

Answer 4

well, sulfur, we have 0.6 moles (20g / 32g/mol = 0.6mol)
O2, we have 5 moles (160g / 32g/mol = 5mol)
Sulfur is our limiting reagent, we cannot make any more than 0.6 moles of SO2

0.6moles(S) * 32g/mol + 0.6moles(O2) * 32g/mol
= 19.2g(S) * 19.2g(O2)
= 38.4g(SO2)

Answer 5

law of conservation of mass- mass cannot be created nor destroyed. So you simply add them together!