Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio do masses of oxygen combine with a fixed mass of manganese in these two compounds

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

 50.48 % O = 50.48 grams

 49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

i do not get it   at all

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2