If a #49.9*g# mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?

Question

Ellie Andrews · Accepted Answer

If we assume #"sodium hydroxide"# reacts with #"sulfuric acid"#.....approx. #90*g# #"sodium sulfate"# results......

We write the equation as: #2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O(l)# #"Moles of sodium hydroxide"=(49.9*g)/(40.00*g*mol^-1)=1.25*mol#. The equation specifies that HALF an equiv of #"sodium sulfate"# results from neutralization, and this represents a mass of #1.25*molxx1/2xx142.04*g*mol^-1=88.8*g.#

Owen Ambrose · Answer

undefined To find the mass of sodium sulfate formed, you need to first balance the chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4). The balanced equation is:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the balanced equation, you can see that 2 moles of NaOH react with 1 mole of H2SO4 to produce 1 mole of Na2SO4.

Calculate the number of moles of NaOH using its molar mass (40 g/mol) and the given mass:

49.9 g / 40 g/mol = 1.2475 moles

According to the stoichiometry of the reaction, 2 moles of NaOH react to form 1 mole of Na2SO4. Therefore, the number of moles of Na2SO4 formed is half of the moles of NaOH:

1.2475 moles NaOH * (1 mole Na2SO4 / 2 moles NaOH) = 0.6238 moles Na2SO4

Now, calculate the mass of Na2SO4 using its molar mass (142 g/mol):

0.6238 moles * 142 g/mol = 88.64 g

So, the mass of sodium sulfate formed is approximately 88.64 grams.

Peyton Adams · Answer

undefined To determine the mass of sodium sulfate produced when 49.9 g of sodium hydroxide reacts with sulfuric acid, we need to consider the balanced chemical equation for the reaction between sodium hydroxide ($ NaOH $) and sulfuric acid ($ H2SO4 $).

The balanced chemical equation for the reaction is:

$$ 2NaOH + H2SO4 \rightarrow Na2SO4 + 2H2O $$

From the equation, we see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid to produce 1 mole of sodium sulfate.

First, we need to determine the number of moles of sodium hydroxide provided:

$$ \text{Number of moles of } NaOH = \frac{\text{Mass}}{\text{Molar mass}} $$

The molar mass of sodium hydroxide ($ NaOH $) is approximately 40 g/mol (22.99 g/mol for sodium + 16.00 g/mol for oxygen + 1.01 g/mol for hydrogen).

$$ \text{Number of moles of } NaOH = \frac{49.9 \text{ g}}{40 \text{ g/mol}} \approx 1.2475 \text{ mol} $$

According to the balanced equation, 2 moles of sodium hydroxide produce 1 mole of sodium sulfate. Therefore, the number of moles of sodium sulfate produced is half the number of moles of sodium hydroxide:

$$ \text{Number of moles of } Na2SO4 = \frac{1}{2} \times 1.2475 \text{ mol} \approx 0.62375 \text{ mol} $$

Now, we can calculate the mass of sodium sulfate produced:

$$ \text{Mass of } Na2SO4 = \text{Number of moles} \times \text{Molar mass} $$

The molar mass of sodium sulfate ($ Na2SO4 $) is approximately 142.04 g/mol (22.99 g/mol for sodium + 32.06 g/mol for sulfur + 4 × 16.00 g/mol for oxygen).

$$ \text{Mass of } Na2SO4 = 0.62375 \text{ mol} \times 142.04 \text{ g/mol} \approx 88.58 \text{ g} $$

Therefore, approximately 88.58 grams of sodium sulfate will result from the reaction.