When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 16.8 g of carbon were burned in the presence of 59.4 g of oxygen, 14.6 g of oxygen remained unreacted. What mass of carbon dioxide was produced?

61.6 g

Explanation:

Step 1: Write the balanced equation

C + O₂ → CO₂

Since there are 14.6 g of oxygen that remain unreacted by the end of the reaction, oxygen is the reactant in excess and carbon is the limiting reactant, thus the one used to calculate the mass of carbon dioxide produced.

Step 2: Calculate the moles corresponding to 16.8 g of C

The molar mass of C is 12.01 g/mol

16.8 g × 1 mol/12.01 g = 1.40 mol

Step 3: Calculate the number of moles of CO₂ produced from 1.40 moles of C

The molar ratio of C to CO₂ is 1:1. The moles of CO₂ produced are 1/1 × 1.40 mol = 1.40 mol.

Step 4: Calculate the mass corresponding to 1.40 moles of CO₂

The molar mass of CO₂ is 44.01 g/mol.

1.40 mol × 44.01 g/mol = 61.6 g

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