Question

Based on the following chemical equation:

4HCN + 5O₂   -> 2N₂ + 4CO₂ + 2H₂O

Identify the limiting reactant and the mass of N₂ produced when 100.0 g of HCN react with 100.0 g of O₂.

Enter the chemical formula of the limiting reactant.

The reactant that is present in excess will be (enter the chemical formula):

The mass of N₂ produced will be  g. (3 sig figs will be sufficient)

QUESTION 17 Based on the following chemical equation: 4HCN 502 > 2N2+ 4CO2 + 2H20 Identify the limiting reactant and the mass of N2 produced when 100.0 g of HCN react with 100.0 g of O2. Enter the chemical formula of the limiting reactant The reactant that is present in excess will be (enter the chemical formula): g. (3 sig figs will be sufficient) The mass of N2 produced will be

Answer 1

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Question 11 - 4HON + 50₂ → 2N₂ + 4002 + 2H2O molar mass 27.0 32.0 28.0 (in g/mol) 4 moles of HON react with 5 moles of O2 4X27.0g of HON react with 5x320 g of O2 loo.og of HON react with us = 5x320g x 100 og 4X27.09 - 148.15 g 02 (Required amount of O2) But we have 100.09 ol Os It meango is limiting reactant and HON is excen reactant. → Hence, the chemical formella By limiting reactant is 02 | The reactant that is exces will be HON As the reaction proceed according to amount of limiting reactant so 5 moles of O₂ produce I moles of Na 5x3210 g of O2 produce 2x 280 g of No 100.0 g of 2 will produce No = 882810g x 100 q. 5X32109 - 35.09 of N2 Hence, the mass of Ne produced will be 35.0 g .