For each reaction, calculate how many moles of the barium product you will produce using stoichiometry and the balanced reactions. since there are two reactants, calculate the moles of product using each reactant, and then use the number of moles which is less) this comes from the limiting reactant).

Moles of Ba(NH2SO3)2.

Explanation:

given that

mass of Ba(NO3)2 = 1.40g

mass of NH2SO3H = 2.50 g

1)to determine the mole of  Ba(NO3)2

2) to determine the mass of all three product formed in the reaction

reaction

Ba(NO3)2 + 2NH2SO3H → Ba(NH2SO3)2 + 2HNO3

Solution

we calculate the molar mass of each species by using their atomic masses

BA = 137.33g/mol

N = 14g/mol

O= 16g/mol

H = 1g/mol

S = 32g/mol

calculation

Ba(NO3)2 = Ba + 2N + 6O

= 137.33 + 2X 14 + 6 X 16

= 261.33g/mol

NH2SO3H = N + 3H + S+ 3O

=14 + 3X1 + 32 + 3X 16

= 97g/mol

Ba(NH2SO3)2 = Ba + 2N + 4H +2S +6O

= 137.33 + 2 X 14 + 4 X1 + 2X32 + 6 X 16

= 329.33g/mol

HNO3 = H + n + 3O

= 1 + 14 + 3 X 16

= 63g/mol

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.