Final answer:
About 0.01951 grams of sodium sulfide are required to react with 25.00 mL of a 0.0100 M solution of cadmium (II) nitrate to form a precipitate, which, when rounded, is close to 1 gram (answer choice a).
Step-by-step explanation:
To determine how many grams of sodium sulfide (Na2S) are needed to react with 25.00 mL of a 0.0100 M solution of cadmium (II) nitrate (Cd(NO3)2) to form a precipitate of cadmium sulfide (CdS), we need to write out and balance the chemical equation and then use stoichiometry to solve for the mass of Na2S.
The balanced chemical equation is:
Cd(NO3)2(aq) + Na2S(aq) → CdS(s) + 2NaNO3(aq)
The reaction shows that one mole of Cd(NO3)2 reacts with one mole of Na2S to form one mole of CdS. First, we need to calculate the moles of Cd(NO3)2 in 25.00 mL of 0.0100 M solution:
Moles of Cd(NO3)2 = volume (L) × molarity (M) = 0.025 L × 0.0100 M = 0.00025 moles
Since the stoichiometry is 1:1, we need 0.00025 moles of Na2S. The molar mass of Na2S is approximately 78.04 g/mol.
Mass of Na2S = moles × molar mass = 0.00025 moles × 78.04 g/mol = 0.01951 g
Thus, 0.01951 grams of sodium sulfide are needed, which is closest to answer choice (a) 1 gram, assuming we are rounding to the nearest gram.