Determine the mass of AgCl(s) produced when 0.91 g of AgNO3 in an aqueous solution reacts with excess NaCl aqueous solution as shown below: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) i Respond with the correct number of significant figures in scientific notation (Use E notation and only 1 digit before decimal e. g. 2.5E5 for 2.5 x 10")​

0.77 g

Explanation:

Step 1: Write the balanced equation

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Step 2: Calculate the moles corresponding to 0.91 g of AgNO₃

The molar mass of AgNO₃ is 169.87 g/mol.

0.91 g × 1 mol/169.87 g = 5.4 × 10⁻³ mol

Step 3: Calculate the moles of AgCl produced from 5.4 × 10⁻³ moles of AgNO₃

The molar ratio of AgNO₃ to AgCl is 1:1. The moles of AgCl produced are 1/1 × 5.4 × 10⁻³ mol = 5.4 × 10⁻³ mol.

Step 4: Calculate the mass corresponding to 5.4 × 10⁻³ moles of AgCl

The molar mass of AgCl is 143.32 g/mol.

5.4 × 10⁻³ mol × 143.32 g/mol = 0.77 g

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )