Answer:
See explanation for detailed solution
Explanation:
The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3
Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles
From the reaction equation;
1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2
5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2
For HSO3NH2
Number of moles = 2.4g/97.10 g/mol =0.0247 moles
2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2
0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles
Hence, Ba(NO3)2 is the limiting reactant
The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g
b)
Number of moles = mass/ molar mass
Molar mass = mass/ number of moles
Molar mass = 1.6925 g/5.36 × 10^-3 moles = 315.76 g
Answer:
1. Theoretical yield = 2.03g
2. Actual yield 1.89g
Explanation:
Let us write a balanced equation. This is illustrated below:
Zn + 2HCI —> ZnCl2 + H2
Molar Mass of HCl = 1 +35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 2 x 36.5 = 73g
Molar Mass of H2 = 2x1 = 2g/mol
1. From the equation,
73g of HCl produced 2g of H2.
Therefore, 74g of HCl will produce = (74 x 2)/73 = 2.03g
Therefore, theoretical yield = 2.03g
2. %yield = 93%
Theoretical yield = 2.03g
Actual yield =?
%yield = Actual yield /Theoretical yield x100
Actual yield = %yield x theoretical yield
Actual yield = 93% x 2.03 = (93/100)x2.03 = 1.89g
Actual yield =1.89g
Answer:
Both densities would be equal.
Explanation:
since both the bar and the ingot are pure gold, they would have the same density, despite the amount.
The mole fraction of HNO3 is 0.225
<u>Explanation:</u>
<u>1.</u>Given data
Density = 1.429 /ml
Mass% = 63.01 g HNO3 / 100g of solution
The mass of 63.01 g is in 100 / 1.142 /ml of solution
Or 63.01 g in 55.7 mL
Molarity = 15.39 moles / L
Mass of water in 100g = 100 - 63.01=36.99 g
So 63.01 grams in 36.99 grams of water
So mass of HNO3 in 1000grams of water = 63.01* x 1000 / 36.99 = 1703
Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles
Molality = 27.03 molal (mole / Kg)
Mole fraction = Mole of HN03 / Moles of water + mole of HNO3
Mole of water = 62/ 18 = 3.44
Moles of HNO3 = 63.01 / 63.01 = 1.000
Mole fraction = 1.000 / 3.44 + 1.000 = 0.225
The mole fraction of HNO3 is 0.225
Balanced equation
Mg₃(PO₄)₂ + 2Al(NO₃)₃ → 3Mg(NO₃)₂ + 2AlPO₄
<h3>Further explanation</h3>
Given
Reaction
Mg3(PO4)2 + Al(NO3)3 → Mg(NO3)2 + AlPO4
Required
Balanced equation
Solution
Give a coefficient :
Mg₃(PO₄)₂ + aAl(NO₃)₃ → bMg(NO₃)₂ + cAlPO₄
Make an equation :
Mg, left = 3, right=b⇒b=3
P, left=2, right=c⇒c=2
Al, left=a, right=c⇒a=c⇒a=2
The reaction becomes :
Mg₃(PO₄)₂ + 2Al(NO₃)₃ → 3Mg(NO₃)₂ + 2AlPO₄
