Question

note: answers for questions 6, 7, 8, 10, and 11 are submitted with blanks on d2l. work totals 8pts when uploaded with signature.

1. if 3.80 moles of nabr were added to the reaction with excess cl₂, how many moles of nacl would be produced? show work with units for full credit. (3 pts total: 2 pts for answer with units; 1 pt for work)
2. if 2.46 g of nabr were added to the reaction with excess cl₂, how many moles of nacl would be produced? show work with units for full credit. (4 pts total: 2 pts for answer with units; 2 pts for work)
3. in a reaction, 0.318 mol of nacl are produced. how many grams of nabr would be needed? (4 pts total: 2 pts for answer with units; 2 pts for work)

Explanation:

Step1: Write the balanced chemical equation

The balanced chemical equation for the reaction is \(2NaBr + Cl_{2}=2NaCl + Br_{2}\). From this equation, the mole - ratio of \(NaBr\) to \(NaCl\) is \(n(NaBr):n(NaCl)=2:2 = 1:1\).

Step2: Use the mole - ratio to find the moles of \(NaBr\)

Given that \(n(NaCl)=0.318\ mol\). Since the mole - ratio \(n(NaBr):n(NaCl)=1:1\), then \(n(NaBr)=n(NaCl)\).

Step3: Calculate the mass of \(NaBr\)

The molar mass of \(NaBr\), \(M(NaBr)=22.99\ g/mol+79.904\ g/mol = 102.894\ g/mol\).
Using the formula \(m = n\times M\), where \(n = 0.318\ mol\) and \(M = 102.894\ g/mol\).
\(m(NaBr)=0.318\ mol\times102.894\ g/mol\)
\(m(NaBr)=0.318\times102.894\ g\approx32.7\ g\)

Answer:

\(32.7\ g\)