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
When finding empirical formulas decimals are fine so you wouldn't have to multiply all of them to turn into whole numbers. What you would do is divide each of those numbers by the smallest of those numbers and take the nearest whole number so if you had:
1.81 mol .6 mol and .3.61 mol you would divide each of those numbers by .6
and you would have 3, 1, and 6
for your problem you would divide them all by 1 and Carbon would be 2 (if your teacher normally has you round down at .5's then it would be 1) hydrogen would be 3 and oxygen would be 1.
C2H3O
Answer:
1.608 g/cm3
Explanation:
Formula:
D=m/v
d= density
m= mass
v= volume
Given data:
Mass= 19.3 g
Volume= 12 cm3
The volume of water displaces by object is 12 cm³ which means that object hold the volume of 12 cm³
Now we will put the values in formula:
D=19.3 g/ 12 cm3= 1.608 g/cm3
so the density of object is 1.608 g/cm3