How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda? Show your work.
1 answer:
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Answer:
15.9 g
Explanation:
(Take the atomic mass of C=12.0, H=1.0, O=16.0)
no. of moles = mass / molar mass
no. of moles of octane used = 11.2 / (12.0x8 + 1x18)
= 0.0982456 mol
Since oxygen is in excess and octane is the limiting reagent, the no. of moles of H2O depends on the no. of moles of octane used.
From the balanced equation, the mole ratio of octane : water = 2:18 = 1: 9,
so this means, one mole of octane produced 9 moles of water.
Using this ratio, we can deduce that (y is the no. of moles of water produced):
y = 0.0982456x9
y= 0.88421 mol
Since mass = no. of moles x molar mass,
mass of water produced = 0.88421 x (1.0x2+16.0)
=15.9 g
Answer:
Part 1
1. empirical formula is = N₂O₃
2. empirical formula is = NaClO₄
3. empirical formula is = BaCr₂O₇
Part2
no. of atoms of P₄ = 2.1 x 10²³
Part 3
A) no. of moles of S = 0.88 moles
B) no. of atoms of Mg = 1.08 x 10²⁴
C) no. of moles of Br₂ = 9.5 mole
Part 4
A) Molar mass of Na₂SO₄ = 142 g/mol
B) Molar mass of Al₂(SO₄)₃ = 342 g/mol
C) Molar mass of Al₂(SO₄)₃ = 176.5 g/mol
D) Molar mass of K₂CrO₄ = 194 g/mol
Part 5,
mass in grams of I₂ = 254 g
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Explanation:
Part 1:
Empirical Formula Calculation from %
1): Data Given
Percent mass of N = 63.6 %
Percent mass of O = 36.4 %
First convert percent to mass
let say we have 100 g of compound
So
mass of N = 63.6 /100 x 100 = 63.6 g
mass of O = 36.4 /100 x 100 = 36.4 g
Now convert masses to moles:
Molar mass of N = 14 g/mol
Molar mass of O = 16 g/mol
Formula used:
no. moles = mass in gram / molar mass ....................(1)
Now find the no. of moles of nitrogen
Put values in formula 1
no. moles = 36.4 g / 14 g/mol
no. moles = 2.6 mol
Now find the no. of moles of Oxygen
Put values in formula 1
no. moles = 63.6 g / 16 g/mol
no. moles = 4 mol
Now calculate the mole ratio of both element
N = 2.6 /2.6 = 1
O = 4 /2.6 = 1.5
To convert the ratio to whole number multiply the ratio with a whole number.
N = 1 x 2 = 2
O = 1.5 x 2 =3
So,
the ratio of N to O 2 : 3 and this is the simplest form
So the empirical formula is = N₂O₃
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2): Data Given
Percent mass of Na = 18.8 %
Percent mass of Cl = 29 %
Percent mass of O = 52.3 %
First convert percent to mass
let say we have 100 g of compound
So
mass of Na = 18.8 /100 x 100 = 18.8 g
mass of Cl = 29 /100 x 100 = 29 g
mass of O = 52.3 /100 x 100 = 52.3 g
Now convert masses to moles:
Molar mass of Na = 23 g/mol
Molar mass of Cl = 35.5 g/mol
Molar mass of O = 16 g/mol
Formula used:
no. moles = mass in gram / molar mass ....................(1)
Now find the no. of moles of Na
Put values in formula 1
no. moles = 18.8 g / 23 g/mol
no. moles = 0.82 mol
Now find the no. of moles of Cl
Put values in formula 1
no. moles = 29 g / 35.5 g/mol
no. moles = 0.82 mol
Now find the no. of moles of Oxygen
Put values in formula 1
no. moles = 52.3 g / 16 g/mol
no. moles = 3.3 mol
Calculate the mole ratio of both element
Na = 0.82 / 0.82 = 1
Cl = 0.82 / 0.82 = 1
O = 3.3 / 0.82 = 4
So,
The ratio of Na, Cl and O is 1 : 1 : 4 and this is the simplest form.
So the empirical formula is = NaClO₄
_________________________________
3): Data Given
Percent mass of Ba = 38.9 %
Percent mass of Cr = 29.4 %
Percent mass of O = 31.7 %
First convert percent to mass
let say we have 100 g of compound
So
mass of Ba = 38.9 /100 x 100 = 38.9 g
mass of Cr = 29.4 /100 x 100 = 29 g
mass of O = 31.7 /100 x 100 = 31.7 g
Now convert masses to moles:
Molar mass of Ba = 137 g/mol
Molar mass of Cr = 52 g/mol
Molar mass of O = 16 g/mol
Formula used:
no. moles = mass in gram / molar mass ....................(1)
Now find the no. of moles of Ba
Put values in formula 1
no. moles = 38.9 g / 137 g/mol
no. moles = 0.28 mol
Now find the no. of moles of Cr
Put values in formula 1
no. moles = 29.4 g / 52 g/mol
no. moles = 0.56 mol
Now find the no. of moles of Oxygen
Put values in formula 1
no. moles = 31.7 g / 16 g/mol
no. moles = 2 mol
Calculate the mole ratio of both element
Ba = 0.28 / 0.28 = 1
Cr = 0.56 / 0.28 = 2
O = 2 / 0.28 = 7
So,
The ratio of Ba, Cr and O is 1 : 2 : 7 and this is the simplest form.
So the empirical formula is = BaCr₂O₇
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