Asked by Arik
Consider the formation of cyanogen c2n2 and its subsequent decomposition in water given by the equations
2Cu2+ + 6CN- 2[Cu(CN)2]- + C2N2
C2N2 + H2O HCN + HOCN
How much hydrocyanic acid HCN can be produced from 35.00 g of KCN, assuming 100% yield?
2Cu2+ + 6CN- 2[Cu(CN)2]- + C2N2
C2N2 + H2O HCN + HOCN
How much hydrocyanic acid HCN can be produced from 35.00 g of KCN, assuming 100% yield?
Answers
Answered by
DrBob222
I think it extremely advantageous for you to find the arrow key on the computer. Arrows separate reactants from products. They are made easily as ==> or --> or >>>.
Convert 35.00 g KCN to mols. mol = grams/molar mass = ? mols KCN
Using the coefficients in the balanced equationS to convert mols KCN to mols HCN. That will require two factors; i.e.,
?mol KCN x (1 mol C2N2/6 mol KCN) x (1 mol C2N2/1 mol HCN) = ?
Now convert mols HCN to grams. g = mols x molar mass.
Convert 35.00 g KCN to mols. mol = grams/molar mass = ? mols KCN
Using the coefficients in the balanced equationS to convert mols KCN to mols HCN. That will require two factors; i.e.,
?mol KCN x (1 mol C2N2/6 mol KCN) x (1 mol C2N2/1 mol HCN) = ?
Now convert mols HCN to grams. g = mols x molar mass.
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