What volume of hydrogen is released by the interaction of 45 grams of sodium in which 5

What volume of hydrogen is released by the interaction of 45 grams of sodium in which 5 percent of impurities with 200 grams of a 10 percent solution of sulfuric acid

Given:

m (Na) = 45 g

w (impurities) = 5% or 0.05 (convert to fractions of a unit, get rid of percentages)

m solution (H2SO4) = 200 g

w (H2SO4) = 10% or 0.1 (convert to fractions of one, get rid of percentages)

v (H2) -?

Solution

1) w (impurities) = m (impurities) * 100% / m (mixtures)

m (impurities) = 45 g * 0.05 = 2.25 g

m (Na) pure = 45 g – 2.25 g = 42.75 g

2) w (H2SO4) = m (H2SO4) * 100% / m solution (H2SO4)

m (H2SO4 = w (H2SO4) * m solution (H2SO4)

w (H2SO4) = 200g * 0.1 = 20g

3) Find the amount of sodium and sulfuric acid:

n = m / M

M (Na) = 23 g / mol

n (Na) = 42.75 g / 23 g / mol = 1.86 mol

M (H2SO4) = 98 g / mol

n (H2SO4) = 20 g / 98 g / mol = 0.2 mol

4) 2 Na + H2SO4 = Na2SO4 + H2

1 mol H2SO4 – 2 mol Na

0.2 mol H2SO4 – 0.4 mol Na, therefore, sodium is in excess and the calculation is done using sulfuric acid.

V m = 22.4 l / mol

98 g (H2SO4) – 22.4 L (H2)

20 g (H2SO4) – xl (H2)

X = 20 g * 22.4 l / 98 g

X = 4.57 l (H2)

Answer: 4.57 L (H2)