## anonymous one year ago 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?

1. Rushwr

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

2. Rushwr

$SO _{2} : 2H _{2}O$

3. Rushwr

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

4. Rushwr

2*

5. Rushwr

$n= \frac{ m }{M _{w}$

6. Rushwr

$n=\frac{ m }{ M _{w}}$

7. Rushwr

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