Question

If Hydrogen Sulfate is reacted with Copper metal, the resultant products are Sulfur Dioxide gas, liquid water, and aqueous Copper(II) Sulfate. How many grams of water will be produced at the same time that 10.0 grams of Sulfur Dioxide is produced?

Answer 1

\[Cu + 2H _{2}SO _{4} \rightarrow SO _{2} + 2H _{2}O + CuSO _{4}\]

Answer 2

\[SO _{2} : 2H _{2}O\]

Answer 3

So when 1 mole of SO2 is formed @ moles of H2O is formed

Answer 4

2*

Answer 5

\[n= \frac{ m }{M _{w} \]

Answer 6

\[n=\frac{ m }{ M _{w}}\]

Answer 7

where n= No. of moles
m= given mass
Mw= relative molecular mass
Hence by using that we can find the no. of moles of SO2 (relative atomic mass of SO2 will be ----> 32+(16*2) = 64
SO the no. of moles will be 10 divided by 64 which is equal to 0.15625
Hence the moles of H2O formed will be the twice of that of SO2 which 0.3125
Now using the same equation we can find the mass of the water formed by multiplying no. of moles by relative atomic mass ( r.m.m. is 18)
So the mass of H2O formed is 5.625g