Question 1 (Limiting Reagent)

12-15. 00 g aluminum sulfide & 15. 00 g water react until the limiting reagent is employed up. [Atomic mass: H = 1 . 008, Al = 26. 98, S sama dengan 32. 07, O = 16. 00]

This can be a balanced equation for the reaction:

Al2S3 + 6 INGESTING WATER ( a couple of Al (OH)3 + a few H2S

(i) Which in the two reactants is the limiting reagent?

(ii) What is the utmost mass of H2S which may be

formed from these reactants?

(iii) How much excess reagent remains following your

reaction is usually complete?

Answers:

(i) Remedy for determining the constraining reagent

Identify the moles of Al2S3 and WATER

aluminum sulfide: 15. 00 g Г· 150. 158 g/mol sama dengan 0. 099895 mol

water: 10. 00 g Г· 18. 015 g/mol sama dengan 0. 555093 mol

Divide each gopher amount by equation agent

aluminum sulfide: 0. 099895 mol Г· 1 mol = zero. 099895

normal water: 0. 555093 mol Г· 6 mol = 0. 0925155 which can be less than zero. 099895 (for aluminum sulfide) Therefore , normal water is the limiting reagent.

(ii) Solution to get mass of H2S produced

Now that we know the limiting reagent is normal water, this problem turns into " Just how much H2S is usually produced from 15. 00 g of WATER and surplus aluminum sulfide? " Decide moles of 10. 00 g of H2O

drinking water: 10. 00 g Г· 18. 015 g/mol = 0. 555093 mol

Make use of molar percentages to determine skin moles of H2S produced from over amount of water. (a) the H2O/H2S ratio can be 6/3, a 2/1 rate.

(b) water is linked to the two.

What this means is the H2S amount can be one-half water value sama dengan 0. 2775465 mol. Convert moles of H2S to grams. 0. 2775465 mol x 34. 0809 g/mol = 9. 459 g (iii) Solution for surplus reagent staying

We will use the amount of water to estimate how much Al2S3 reacts, then simply subtract that amount from 12-15. 00 g. Determine skin moles of 10. 00 g of H2O: water: twelve. 00 g Г· 18. 015 g/mol = 0. 555093 mol Use large molar ratios to determine moles of Al2S3 that reacts with all the above quantity of water. (a) the Al2S3/H2O proportion is 1/6

(b) normal water is linked to the 6. What this means is the Al2S3 amount is usually one-sixth the value = 0. 09251447...