Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry ) 
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
1) Write the balanced chemical equation
2HCl + Na2 CO3 ----------> 2NaCl + H2CO3
2) Write the molar ratios:
2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3
3) Convert 0.15g of sodium carbonate to number of moles
3a) Calculate the molar mass of Na2CO3
Na: 2 * 23 g/mol = 46 g/mol
C: 12 g/mol =
O: 3 * 16 g/mol = 48 g/mol
molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol
3b.- Calculate the number of moles of Na2CO3
# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3
4) Calculate the number of moles of HCl from the molar proportion:
[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl
5) Calculate the volume of HCl from the definition of Molarity
Molarity, M = # moles / volume in liters
=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters
0.028 liters * 1000 ml / liter = 28 ml.
Answer: 28 mililiters of 0.1 M HCl.
Answer:
2 moles
Explanation:
The following were obtained from the question:
Molarity = 0.25 M
Volume = 8L
Mole =?
Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:
Molarity = mole of solute/Volume of solution.
With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:
Molarity = mole of solute/Volume of solution.
0.25 = mole of MgCl2 /8
Cross multiply to express in linear form
Mole of MgCl2 = 0.25 x 8
Mole of MgCl2 = 2 moles
Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution
Answer:
2,75 mol of O2 it's 88 g of O2.
Explanation:
The weight of the diatomic molecule O2 is 32 g/mol. So considering that, you should multiply 2,75 mol · 32 = 88g :)
