A 2-g sample of sucrose is 6.50% hydrogen. What is the mass percentage of hydrogen in 300 g of sucrose? Show all work to support
1 answer:
The mass percentage of hydrogen in 300 g of sucrose is 9.75%
From the question given above, we were told that:
2 g of sucrose contains 6.50% (ie 0.065) hydrogen
Thus, we can obtain the percentage of hydrogen in 300 g of sucrose as follow:
2 g of sucrose contains 0.065 hydrogen.
Therefore,
300 g of sucrose will contain = = 9.75% of Hydrogen.
Thus, 300 g of sucrose contain 9.75% of Hydrogen.
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Answer:
1. 1 × 10⁻²
Explanation:
Given that:
Temperature = 335 ° C = (335+ 273)K = 608
Pressure = 0.750 atm
Volume = 1 Litre
number of moles of NO2 = ???
Rate Constant =0.0821 L atm /K/mol
Using the Ideal gas equation
PV = nRT
n =
n =
n = 0.015
n = 1.5 × 10⁻² mole
Density = 0.525 g/L
The equation for the reaction can be illustrated as:
2NO2(g) ⇌ 2NO(g) + O2(g)
For the ICE table; we have:
Initial x 0 0
Change -2y + 2y +y
Equilibrium (x - 2y) 2y y
Total moles at equilibrium = (x-2y)+2y+y
= x + y moles
However,
1.5 × 10⁻² mole of the mixture has a mass of 0.525 g
i.e x + y moles = 1.5 × 10⁻² mole
Now, molar mass of 1 mole of NO2 = 46g/mol
Since number of moles =
mass of (x-2y) moles = 46 × (x-2y) g
Molar mass of NO = 30 g/mol
Also, mass of NO = 2y × 30 = 60y
Molar mass of O2 = 32 g/mol
Mass of O2 = y × 32 = 32y
Total mass = ( 46x - 90y)+60y+32y = 0.525
46x = 0.525
x =
x = 0.0114
x = 1.14 × 10⁻²
x + y moles = 1.5 × 10⁻²
y = 1.5 × 10⁻² - 1.14 × 10⁻²
y = 0.0036
y = 3.6 × 10⁻³
At equilibrium
[NO2] = ( 1.14 - 2(0.36))× 10⁻² = 4.2 × 10⁻³ M
[NO] = 2 ( 3.6 × 10⁻³) = 7.2 × 10⁻³ M
[O2] = 3.6 × 10⁻³ M
0.011
1. 1 × 10⁻²