Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
Answer:
26.67 mol HCl
Explanation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.
To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:
- 8.89 mol Al(OH)₃ *
= 26.67 mol HCl
Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
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Molarity of Ag+ is less than the molar solubility thus ppt will not occur.
Balanced reaction-:
<h3>2AgNO3(aq)+K2CrO4(aq)→Ag2CrO4(s)+2KNO3(aq)</h3>
Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol
=1.589⋅10^−7 mol
Molarity of Ag+=moles of solute(L)=1.589⋅10−7 mol0.015 L=1.059⋅10−5M
Ksp of Ag2CrO4
=[Ag+]2[CrO42−]
1.2⋅10−12=[2s]2[s]
4s3=1.2⋅10−12
s=6.69⋅10−5 M
Molarity of Ag+ is less than the molar solubility thus ppt will not occur.
<h3>What is the molarity calculation formula?</h3>
The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.
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