Respuesta :
Answer: A. 1.5 g
Explanation:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of nitrogen}=\frac{7.0g}{28g/mol}=0.25 moles[/tex]
According to stoichiometry:
1 mole of [tex]N_2[/tex] requires = 3 moles of [tex]H_2[/tex]
Thus 0.25 moles of [tex]N_2[/tex] will require =[tex]\frac{3}{1}\times 0.25=0.75moles[/tex] of [tex]H_2[/tex]
Mass of [tex]H_2[/tex] required =[tex]moles\times {\text {Molar mass}}=0.75mol\times 2g/mol=1.5g[/tex]
The minimum amount of [tex]H_2[/tex] in grams that would be required to completely react with this amount of [tex]N_2[/tex] is 1.5 grams.
Answer:
The correct answer is option A.
Explanation:
[tex]N_2 + 3 H_2\rightarrow 2 NH_3[/tex]
Moles of nitrogen gas = [tex]\frac{7.0 g}{28 g/mol}=0.25 mol[/tex]
According to reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen gas.
Then 0.25 moles of nitrogen gas will react with:
[tex]\frac{3}{1}\times 0.25 mol=0.75 mol[/tex] of hydrogen gas.
Mass of 0.75 moles of hydrogen gas = 0.75 mol × 2 g/mol = 1.5 g
1.5 grams of hydrogen that would be required to completely react with this amount of nitrogen.