Answer:
There will be produced:
2.97 moles HMnO4
4.45 moles Pb(NO3)2
2.97 moles H2O
Explanation:
Step 1: Data given
Manganese(II) oxide = MnO2
lead(IV) oxide = PbO2
nitric acid = HNO3
Moles of HNO3 = 8.90 moles
Step 2: The balanced equation
2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb(NO3)2 + 2H2O
Step 3: Calculate moles of reactants and products
For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb(NO3)2 and 2 moles of water
For 8.90 moles of HNO3, there will react:
8.90 / 3 = 2.97 moles MnO2
8.90 / 2 = 4.45 moles PbO2
There will be produced:
8.90/3 = 2.97 moles HMnO4
8.90/2 = 4.45 moles Pb(NO3)2
8.90 / 3 = 2.97 moles H2O
Use the formula stated below
So
#H_3PO_4
#Ba(OH)_2
Hence H_3PO_4 is the limiting reagent
Combustion of hexane can be illustrated by the following reaction:
2C6H14 + 19O2 ...........> 12CO2 + 14H2O
From the periodic table:
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of carbon = 12 grams
Therefore:
molar mass of hexane = 6(12)+14(1) = 86 grams
mass of water = 2(1)+16 = 18 grams
From the balanced equation above:
2(86) = 172 grams of hexane produce 252 grams of water. To know the amount of water produced from 1.33 grams of hexane, all you have to do is cross multiplication as follows:
amount of water = (1.33x252) / 172 = 1.9486 grams
