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)

Explanation:

Step1: Write the balanced chemical equation

The reaction between \(NaBr\) and \(Cl_2\) is \(2NaBr + Cl_2=2NaCl + Br_2\). From the equation, the mole ratio of \(NaBr\) to \(NaCl\) is \(2:2 = 1:1\).

Step2: Calculate the moles of \(NaCl\) for question 6

Given \(n(NaBr)=3.80\space mol\). Since the mole ratio \(n(NaBr):n(NaCl)=1:1\), then \(n(NaCl)=3.80\space mol\)

Step3: Calculate the molar mass of \(NaBr\) for question 7

The molar mass of \(NaBr\), \(M(NaBr)=M(Na)+M(Br)=22.99\space g/mol + 79.90\space g/mol=102.89\space g/mol\)

Step4: Calculate the moles of \(NaBr\) for question 7

Using the formula \(n=\frac{m}{M}\), where \(m = 2.46\space g\) and \(M = 102.89\space g/mol\). So \(n(NaBr)=\frac{2.46\space g}{102.89\space g/mol}\approx0.0239\space mol\)

Step5: Calculate the moles of \(NaCl\) for question 7

Since the mole ratio \(n(NaBr):n(NaCl)=1:1\), then \(n(NaCl)= 0.0239\space mol\)

Answer:

Question 6: \(3.80\space mol\)
Question 7: \(0.0239\space mol\)