<span>4.3065 g
First, lookup the atomic weights of all the elements involved.
Atomic weight of Calcium = 40.078
Atomic weight of Carbon = 12.0107
Atomic weight of Hydrogen = 1.00794
Atomic weight of Oxygen = 15.999
Atomic weight of Sulfur = 32.065
Now calculate the molar masses of the reactants and product
Molar mass H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999
= 98.07688 g/mol
Molar mass CaCO3 = 40.078 + 12.0107 + 3 * 15.999
= 100.0857 g/mol
Molar mass CaSO4 = 40.078 + 32.065 + 4 * 15.999
= 136.139 g/mol
The balanced reaction for H2SO4 with CaCO3 is
CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2
So it takes 1 mole each of CaCO3 and H2SO4 to produce 1 mole of CaSO4. Let's see how many moles of CaCO3 and H2SO4 we have.
CaCO3: 3.1660 g / 100.0857 g/mol = 0.031632891 mol
H2SO4: 3.2900g / 98.07688 g/mol = 0.033545113 mol
We have a slight excess of H2SO4, so the amount of CaCO3 is the limiting reactant and we should have 0.031632891 moles of product. To determine its mass, multiply the number of moles by the molar mass computed earlier.
0.031632891 mol * 136.139 g/mol = 4.306470148 g
Since we have 5 significant figures in our data, round the result to 5 figures, giving 4.3065 g</span>
Bromine because they are both in Group 17, which means they have the same number of valence electrons
the answer is twenty eight