For the reaction H₂SO₄ + 2NaC≡N → 2HC≡N + Na₂SO₄, we have that 8.4 moles of sodium cyanide are needed to produce 4.2 moles of sodium sulfate (option C).
The reaction between sodium cyanide and sulfuric acid is:
H₂SO₄ + 2NaC≡N → 2HC≡N + Na₂SO₄
We can find the number of moles of sodium cyanide, assuming that it is the limiting reactant and knowing that 2 moles of it produce 1 mol of sodium sulfate, as follows:
[tex] n_{NaC\equiv N} = \frac{2\: moles\: NaC\equiv N}{1 \: mol\: Na_{2}SO_{4}}*4.2 \: moles\: Na_{2}SO_{4} = 8.4 \: moles [/tex]
Therefore, 8.4 moles of sodium cyanide are needed to produce 4.2 moles of sodium sulfate (option C).
Learn more about limiting reactants here:
brainly.com/question/14225536
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