4.8g of calcium is added to 3.6g of water. the following reaction occurs ca + 2h2o > ca(oh)2 + h2 the number of moles of calcium the number of moles of water

Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol 
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation 
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present 

q2)
next we are asked to calculate the number of moles of water present 
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present 

sum of the products of the molar masses of the individual elements by the number of atoms 
H - 1 g/mol and O - 16 g/mol 
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol 
molar mass of H₂O is 18 g/mol 
therefore number of moles of water  = 3.6 g / 18 g/mol = 0.2 mol 
0.2 mol of water is present 

The answer is b metal denting