In the reaction below, how many moles of chlorine are needed to react with 23.9 g of sodium hydroxide?
2Cl2 + NaOH -> 3NaCL + NaClO2 + 2H2O
a. 1.20 mol chlorine
b. 0.598 mol chlorine
c. 0.398 mol chlorine
d. 0.149 mol chlorine
e. 0.299 mol chlorine
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To answer this question, we need to use the molar mass of sodium hydroxide (NaOH) as a conversion factor to convert grams of sodium hydroxide to moles of sodium hydroxide.
The molar mass of sodium hydroxide (NaOH) is:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Adding up the molar masses:
Na + O + H = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Now, we can use the molar mass of sodium hydroxide to convert grams of sodium hydroxide to moles of sodium hydroxide:
23.9 g NaOH x (1 mol NaOH/ 40.00 g NaOH) = 0.598 mol NaOH
According to the balanced equation, the ratio of moles of chlorine to moles of sodium hydroxide is 2:1. Therefore, we need half the amount of moles of chlorine.
0.598 mol NaOH x (1 mol Cl2/ 2 mol NaOH) = 0.299 mol Cl2
Therefore, the answer is e. 0.299 mol chlorine.
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