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Upon balancing the equation below, how many moles of sodium cyanide are needed to react completely with 3.6 moles of sulfuric acid? H2SO4 + NaCN yields HCN + Na2SO4

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You would need 7.2 moles of sodium cyanide to completely react with 3.6 moles of sulfuric acid

Explanation:

The given chemical equation is as follows.

    [tex]H_{2}SO_{4} + NaCN \rightarrow HCN + Na_{2}SO_{4}[/tex]

Number of reactant atoms are as follows.

  • H = 2
  • [tex]SO_{4}[/tex] = 1
  • Na = 1
  • CN = 1

Number of product atoms are as follows.

  • H = 1
  • [tex]SO_{4}[/tex] = 1
  • Na = 1
  • CN = 1

Thus, in order to balance the equation, multiply NaCN by 2 on the reactant side and multiply HCN by 2 on the product side. Therefore, the balanced chemical equation will be as follows.

      [tex]H_{2}SO_{4} +2NaCN \rightarrow 2HCN + Na_{2}SO_{4}[/tex]

Hence, we need [tex]2 \times 3.6[/tex]  = 7.2 moles of NaCN to react completely with 3.6 moles of [tex]H_{2}SO_{4}[/tex].