Answer:
Explanation:
2 H₂S(g) +S0₂(g) = 3 S(s) + 2H₂0(g)
2 x 34 g 64 g 3 x 32 g
68 g of H₂S reacts with 64 g of S0₂
3.89 g of H₂S reacts with 64 x 3 .89 / 68 g of S0₂
3.89 g of H₂S reacts with 3.66 g of S0₂
S0₂ given is 4.11 g , so it is in excess .
Hence H₂S is limiting reagent .
68 g of H₂S reacts with S0₂ to give 96 g of Sulphur
3.89 g of H₂S reacts with S0₂ to give 96 x 3.89 / 68 g of Sulphur
3.89 g of H₂S reacts with S0₂ to give 96 x 3.89 / 68 g of Sulphur
5.49 g of Sulphur is produced .
Actual yield is 4.89
percentage yield = 4.89 x 100 / 5.49
= 89 % .
Answer:
5.0 moles of water per one mole of anhydrate
Explanation:
To solve this question we must find the moles of the anhydrate. The difference in mass between the dry and the anhydrate gives the mass of water. Thus, we can find the moles of water and the moles of water per mole of anhydrate:
<em>Moles Anhydrate:</em>
7.58g * (1mol / 84.32g) = 0.0899 moles XCO3
<em>Moles water:</em>
15.67g - 7.58g = 8.09g * (1mol / 18.01g) = 0.449 moles H2O
Moles of water per mole of anhydrate:
0.449 moles H2O / 0.0899 moles XCO3 =
5.0 moles of water per one mole of anhydrate
Explanation:
2Li+2H2O—>2LiOH+H2
Calculate the mass of reacted lithium when H2 is 6.02 * 10 ^ 23 molecules.
I really need the answer with all the calculation please.
What you were given is the balanced chemical equation
