Question

for this given reaction 2 naoh + cl2 --> naocl + nacl + h2o, 80g of naoh will give
a 18 g of water is formed
b 40 g of water is formed
c 1 g of water is formed
question 2 of 6

Explanation:

Step1: Calculate molar - mass of NaOH and H₂O

The molar - mass of NaOH ($M_{NaOH}$) is $M_{NaOH}=23 + 16+1=40$ g/mol. The molar - mass of H₂O ($M_{H2O}$) is $M_{H2O}=2\times1 + 16 = 18$ g/mol.

Step2: Determine the mole of NaOH

The number of moles of NaOH ($n_{NaOH}$) is calculated using the formula $n=\frac{m}{M}$. Given $m_{NaOH}=80$ g, then $n_{NaOH}=\frac{80}{40}=2$ mol.

Step3: Use the stoichiometric ratio

From the balanced chemical equation $2NaOH + Cl_2
ightarrow NaOCl+NaCl + H_2O$, the mole ratio of NaOH to H₂O is 2:1. So, when $n_{NaOH} = 2$ mol, the number of moles of H₂O formed ($n_{H2O}$) is 1 mol.

Step4: Calculate the mass of water

Using the formula $m = n\times M$, with $n_{H2O}=1$ mol and $M_{H2O}=18$ g/mol, we get $m_{H2O}=1\times18 = 18$ g.

Answer:

A. 18 g of water is formed