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irina1246 [14]
3 years ago
12

Sulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning

it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is
Chemistry
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

The mass of NaHCO3 required is 235.22 g

Explanation:

*******

Continuation of Question:

2NaHCO3(s) + H2SO4(aq)  →  Na2SO4(aq) + 2CO2(g) + 2H2O(l)

You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?

********\

The question requires us to calculate the mass of NaHCO3  to neutralize the acid.

From the balanced chemical equation;

1 mol of H2SO4 requires 2 mol of NaHCO3

1.4 would require x?

Upon solving for x we have;

x = 1.4 * 2 = 2.8 mol of NaHCO3

The relationship between mass and number of moles is given as;

Mass = Number of moles * Molar mass

Mass = 2.8 mol * 84.007 g/mol

Mass =  235.22 g

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2 years ago
A gas sample occupies 3.25 liters at 297.5K and 2.4 atm. Determine the temperature at which the gas will occupy 4.25 L at 1.50 a
lorasvet [3.4K]

For equal moles of  gas, temperature can be calculated from ideal gas equation as follows:

P×V=n×R×T ...... (1)

Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.

2.4 atm ×3.25 L=n×R×297.5 K

Rearranging,

n\times R=0.0262 atm L/K

Similarly at final pressure and volume from equation (1),

1.5 atm ×4.25 L=n×R×T

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Explanation:

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1 How many moles of solute are in:
Sonbull [250]

Moles of solute for both a and b are the same = 1 mol

<h3>Further explanation</h3>

Given

a 500 cm³ of solution, of concentration 2 mol/dm³

b 2 litres of solution, of concentration 0.5 mol/dm³

Required

moles of solute

Solution

Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution

Can be formulated :

\large \boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

a.

V = 500 cm³ = 0.5 L

M = 2 mol/L

n=moles = M x V

n = 2 mol/L x 0.5 L

n = 1 mol

b.  

V = 2 L

M = 0.5 mol/L

n=moles = M x V

n = 0.5 mol/L x 2 L

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3 years ago
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