if 10.0 g of nitrogen undergoes reaction with 10.0 g of hydrogen?
A. 20 g
B. 10 g
C. 12.1 g
D. 56.7 g
5. need a balanced equation
A. 20 g
B. 10 g
C. 12.1 g
D. 56.7 g
5. need a balanced equation
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need a balanced equation
N2 + 3H2 --> 2NH3
10g / 28g/mole = 0.357moles N2
10g / 2g/mole = 5moles H2
5moles H2 x (1N2 /3H2) = moles N2 based upon the balanced equation or 1.67moles N2 required
we only have 0.357moles N2 so N2 = limiting reactant
0.357moles N2 x (2NH3 / 1N2) x 17g/mole = 12.1g
N2 + 3H2 --> 2NH3
10g / 28g/mole = 0.357moles N2
10g / 2g/mole = 5moles H2
5moles H2 x (1N2 /3H2) = moles N2 based upon the balanced equation or 1.67moles N2 required
we only have 0.357moles N2 so N2 = limiting reactant
0.357moles N2 x (2NH3 / 1N2) x 17g/mole = 12.1g