Answer to Question #251357 in General Chemistry for Shy

Question

Show your solution and provide units.

1. Calculate the volume (in mL) of 2.25 M HNO3 solution required to react with 24.2 mL of 3.50 M NaOH solution.

2.What volume (in mL) of 4.35 M HCl solution is needed to react with 18.5 g of CA(OH)2 ?

3.How many grams of silver chromate (Ag2CrO4) will precipitate when 86.0 mL of 0.200 M silver nitrate (AgNO3) is added to 50.0 mL of 0.300 M chromate (K2CrO4)? ( Molar mass of Ag2CrO4 = 331.74 g/ mol).

Answer 1

1. HNO₃ + NaOH → NaNO₃ + H2O

Proportion:

3.50 mol – 1000 mL

x mol – 24.2 mL

x = n(NaOH) $= \frac{3.50 \times 24.2}{1000} = 0.0847 \;mol$

According to the reaction equation:

n(HNO₃) = n(NaOH) = 0.0847 mol

Proportion:

4.35 mol – 1000 mL

0.0847 – y mL

V(HNO₃) $= \frac{0.0847 \times 1000}{4.35} = 19.47\;mL$

2. 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

M(Ca(OH)₂) = 74.09 g/mol

n(Ca(OH)₂) $= \frac{18.5}{74.09} = 0.249 \;mol$

According to the reaction equation:

n(HCl) = 2n(Ca(OH)₂) $= 2 \times 0.249 = 0.498 \;mol$

Proportion:

4.35 mol – 1000 mL

0.498 mol – x mL

x = V(HCl) $= \frac{0.498 \times 1000}{4.35} = 114.5 \;mL$

3. 2AgNO₃ + K₂CrO₄ → Ag₂CrO₄ + 2KNO₃

n(AgNO₃) $= \frac{0.200 \times 86.0}{1000} = 0.0172 \;mol$

n(K₂CrO₄) $= \frac{0.300 \times 50.0}{1000} = 0.015 \;mol$

Accordinf to the reaction for one mole of K₂CrO₄ we need two moles of AgNO_3.

So, AgNO₃ is the limiting reactant.

n(Ag₂CrO₄) = $\frac{1}{2}$ n(AgNO₃) $= \frac{1}{2} \times 0.0172 = 0.0086 \;mol$

m(Ag₂CrO₄) $= 0.0086 \times 331.74 = 2.85 \;mol$