Chemistry math help

Question

What concentration of NO-3 results when  893 mL of 0.509 M NaNO3 is mixed with 793 mL of 0.695 M Ca(NO3)2

Answer 1

NaNO₃ ==> Na⁺ + NO₃^-

0.893 L x 0.509 mol/L x 1 mol NO₃/mole NaNO₃ = 0.455 moles NO₃^-

Ca(NO₃)₂ ==> Ca²⁺ + 2NO₃^-

0.793 L x 0.695 mol/L x 2 mol NO₃/mol CaNO₃ = 1.10 moles NO₃^-

Total moles NO₃^- = 0.455 + 1.10 = 1.56 moles NO₃^-

Total volume in liters = 0.893 + 0.793 = 1.69 liters

[NO₃^-] = 1.56 moles/1.69 L = 0.923 M