Question

1. How many moles of MgS2O3 are in 225 g of the compound?

2. Determine the empirical formula for a compound that is 64.8% C, 13.6% H, and 21.6% O by mass.

3.A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula?

4.The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water.

What is the maximum mass of H2O that can be produced by combining 73.8 g of each reactant?

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

5.The combustion of propane may be described by the chemical equation

C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)

How many grams of O2(g) are needed to completely burn 91.9 g C3H8(g)?

6.Assuming an efficiency of 28.40%, calculate the actual yield of magnesium nitrate formed from 147.3 g of magnesium and excess copper(II) nitrate.

Mg+Cu(NO3)2⟶Mg(NO3)2+Cu

7. If 3.23 mol of C5H12 reacts with excess O2, how many moles of CO2 will be produced by the following combustion reaction?

C5H12+8O2⟶6H2O+5CO2

Answer 1

1)

Molar mass of MgS2O3,

MM = 1*MM(Mg) + 2*MM(S) + 3*MM(O)

= 1*24.31 + 2*32.07 + 3*16.0

= 136.45 g/mol

mass(MgS2O3)= 225 g

use:

number of mol of MgS2O3,

n = mass of MgS2O3/molar mass of MgS2O3

=(2.25*10^2 g)/(1.364*10^2 g/mol)

= 1.649 mol

Answer: 1.65 mol

2)

Let total mass be 100 g

we have mass of each elements as:

C: 64.8 g

H: 13.6 g

O: 21.6 g

Divide by molar mass to get number of moles of each:

C: 64.8/12.01 = 5.3955

H: 13.6/1.008 = 13.4921

O: 21.6/16.0 = 1.35

Divide by smallest to get simplest whole number ratio:

C: 5.3955/1.35 = 4

H: 13.4921/1.35 = 10

O: 1.35/1.35 = 1

  

So empirical formula is:C4H10O

Answer: C4H10O

Only 1 question at a time please. I have answered 2 of them