given the following chemical reaction, calculate the mass of sodium sulfate produced when 9.71 moles of…

given the following chemical reaction, calculate the mass of sodium sulfate produced when 9.71 moles of sulfuric acid (h₂so₄) reacts with excess sodium chloride according to the following balanced reaction: 2 nacl(aq) + 1 h₂so₄(aq) → 1 na₂so₄(aq) + 2 hcl(aq)

given the following chemical reaction, calculate the mass of sodium sulfate produced when 9.71 moles of sulfuric acid (h₂so₄) reacts with excess sodium chloride according to the following balanced reaction: 2 nacl(aq) + 1 h₂so₄(aq) → 1 na₂so₄(aq) + 2 hcl(aq)

Answer

Explanation:

Step1: Determine mole - ratio

From the balanced equation $2NaCl(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq) + 2HCl(aq)$, the mole - ratio of $H_2SO_4$ to $Na_2SO_4$ is $1:1$. So, if 9.71 moles of $H_2SO_4$ react, 9.71 moles of $Na_2SO_4$ are produced.

Step2: Calculate molar mass of $Na_2SO_4$

The molar mass of $Na$ is approximately 22.99 g/mol, $S$ is approximately 32.07 g/mol, and $O$ is approximately 16.00 g/mol. $M(Na_2SO_4)=2\times22.99 + 32.07+4\times16.00=45.98 + 32.07+64.00 = 142.05$ g/mol.

Step3: Calculate mass of $Na_2SO_4$

We use the formula $m = n\times M$, where $n$ is the number of moles and $M$ is the molar mass. $m(Na_2SO_4)=9.71\ mol\times142.05\ g/mol = 1379.3055\ g\approx1379\ g$.

Answer:

1379 g