Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
How much magnesium sulphide can be obtained from 2 g of magnesium and 2 g of sulphur by the reaction:Mg+S→MgS?Which is the limiting reactant?Calculate the amount of one of the reactant whh remains unreacted.
Divide the quantities of both the reactants with their respective gram atomic weights.
Then We have 2/24 = 1/12 = 0.08 mols of Magnesium and 2/32 =1/16 = 0.0625 mols of sulphur .
Then sulphur being in lesser quantities will act as the limiting reagent.
Thus only 0.0625 mols of MgS will be produced having mass 3.5 g and remaing 0.0175 mols of Mg having mass 0.5g
mg is limiting reactant
Mg + S = MgS
Limiting reagent is sulphur as its molar mass is higher.
32g of sulphur reacts with 24g Mg
1g of suphur reacts with 24/32g of Mg
Hence, 2g of sulphur reacts with 24/32*2 g of Mg = 1.5 g
Hence 0.5 g of Mg is unreacted.
PLZZZZZZZZZZZZZ APPROVE IF IT HELPED YOU
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !