Answer to Question #127012 in General Chemistry for Rosé A

Question

1. 1 g of Magnesium is burned in a closed vessel which contains 0.5 g of oxygen.

a. Which is the limiting reagent?

b. What is the amount of magnesium oxide formed in the reaction?

2. How much MgS can be obtained from 2 g of Magnesium and 2 g of sulphur by the reaction , Mg + S→MgS

   a. Which is the limiting reagent?

   b. What is the amount of the unreacted reactant?

3. How many grams of oxygen are required to completely react with 0.2 g of hydrogen to yield water? Also calculate the amount of water formed?

Answer 1

2Mg + O2 = 2MgO

Calculate moles in each case

Moles = mass/molar mass

Moles of Mg = 1/24 = 0.0417

Moles of O2 = 0/5/32 = 0.0156

moles of MgO = moles of a single O and MG

Moles of 0.5/16 = 0.03125 moles

0.03125 + 0. 0417 = 0.07295

mass = molar mass * moles = 0.0795 * (24+16)

mass of MgO formed = 2.918g

2 a) The limiting reagent is sulphur

b )Amount that can formed

 Mg + S→MgS

Moles of Mg = 2/24 = 0.0833g moles

moles of sulphur = 2/32 = 0.0625 moles

Moles = 0.1485

mass = moles *molar mass

0.1485 * 56 = 8.165g

Moles ratio = 0.0833/ 0.0625

1:1.3

the aspect means that 1 mole of suphur reacts to give MgS

thus mass will be 0.0625 * 56 (molar mass of MgS) = 3.25g

mass of the unreacted = 0.0833 - 0.0625 = 0.0208g

mass = 0.0208 * 24( molar mass of unreacted Mg) = 0.04999

which is approximately 0.5g

3) balanced equation

2H_{2 +} O₂ - 2H₂O

moles of

ratio 2:1=2

2mol:1 mol;2mol

4g:32g:36 g

4g of hydrogen requires 32g of Oxygen

0.2g requires ?

cross multiply

0.2*32/4 = 1.6g

1.6g of water is fromed