2 SO2(g) + O2(g) + 2 H2O(ℓ) −→ 2 H2SO4(ℓ) What mass in grams of SO2 is needed to
react with 1527 g of O2?

6116g

Explanation:

2SO2(g) + O2(g) + 2H2O(ℓ) −→ 2H2SO4(ℓ)

We want to find the mass in grams of SO2 that is needed to react with 1527 g of O2. First we must convert the grams of O2 to moles of O2 then to moles of SO2 and then to grams of SO2

So first lets find the molar mass of O2

The mass of oxygen according to a periodic table is 15.999

Using this the mass of O2 would be 15.999(2) = 31.988g

Next we need to identify the mole ratio of O2 to SO2

Looking at the equation for 1 mole of O2 there are two moles of SO2

Next we need to find the molar mass of SO2

Again the mass of oxygen is 15.999g and the mass of Sulfur is 32.066

So the mass of SO2 would be 15.999(2) + 32.066 = 64.064g

Now that we have found all the needed conversions :

We can now use dimensional analysis to calculate the answer.

Kindly check the attached image to see the table. ( sorry if its a bit blurry )
Explanation : The conversions are used to cancel out the units to get to the final unit which is gSO2.

Once the units are cancelled out except for the gSO2 we mutliply and divide based off of what the table says to do.
Here first we divide 1527 by 31.988. We than multiply by 2. Finally we multiply by 64.064 to get the final answer which is 6116gSO2

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3