Question

what products are formed when sodium reacts with gaseous chlorine? select all products formed.
nacl
h₂
na
cl₂
h₂o
question 4
what mass of sodium carbonate would you need to prepare a 25.00 ml solution of 2.57 m na₂co₃? report your answer to 3 sig figs and in grams, but do not include units in the answer.

Explanation:

Step1: Calculate the number of moles of \(Na_2CO_3\)

Use the formula \(n = C\times V\). Given \(C = 2.57\space M\) and \(V=25.00\space mL=25.00\times10^{- 3}\space L\).
\(n = 2.57\times25.00\times10^{-3}=0.06425\space mol\)

Step2: Calculate the molar mass of \(Na_2CO_3\)

The molar mass of \(Na\) is \(22.99\space g/mol\), \(C\) is \(12.01\space g/mol\), and \(O\) is \(16.00\space g/mol\).
\(M = 2\times22.99+12.01 + 3\times16.00=105.99\space g/mol\)

Step3: Calculate the mass of \(Na_2CO_3\)

Use the formula \(m=n\times M\).
\(m=0.06425\times105.99 = 6.81\space g\)

Answer:

6.81