How many grams of #H_2# are needed to produce 11.27 g of #NH_3#?

Answer 1

We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.

#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#

#11.27# #g# of ammonia represents #(11.27*g)/(17.03*g*mol^-1)# #=# #??# #mol#.

How much dinitrogen gas was needed? Regardless of this molar quantity, the reaction's stoichiometry makes it evident that 3/2 equiv of dihydrogen gas were needed.

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Answer 2

Naomi Aronson

2.001 g hydrogen gas is needed to produce 11.27 g ammonia if nitrogen gas is present in excess.

Create a balanced equation first.

#"3H"_2+"N"_2##rarr##"2NH"_3"#

Since the inquiry focuses on the required mass of hydrogen gas, it is assumed that there is an excess of nitrogen gas.

The molar masses of #"H"_2"# and #"NH"_3"# are needed.

#"H"_2":# #"2.01588 g/mol"# http://pubchem.ncbi.nlm.nih.gov/compound/783

#"NH"_3"# #"17.03052 g/mol"# http://pubchem.ncbi.nlm.nih.gov/compound/222#section=Top

We will be making the following conversions: mass #"NH"_3"##rarr##"mol NH"_3"##rarr##"mol H"_2"##rarr##"mass H"_2"#

Determine moles of #"NH"_3"# by dividing the given mass by its molar mass.

Determine moles of #"H"_2"# by multiplying the mole ratio from the balanced equation between #"H"_2"# and #"NH"_3"# with moles #"H"_2"# in the numerator. #(3"mol H"_2)/(2"mol NH"_3)#

Determine mass of #"H"_2"# by multiplying moles #"H"_2"# by its molar mass.

#11.27"g NH"_3xx(1"mol NH"_3)/(17.03052"g NH"_3)xx(3"mol H"_2)/(2"mol NH"_3)xx(2.01588"g H"_2)/(1"mol H"_2)="2.001 g H"_2"# (rounded to four significant figures)

If nitrogen gas is present in excess, 2.001 g of hydrogen gas is required to produce 11.27 g of ammonia.

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Answer 3

Cooper Anderson

To determine the grams of $H_2$ needed to produce 11.27 g of $NH_3$ , we use stoichiometry.

Write the balanced chemical equation: $N_2 + 3H_2 \rightarrow 2NH_3$
Calculate the molar mass of $NH_3$ : $Molar\ mass\ of\ NH_3 = 14.01\ g/mol + 3(1.01\ g/mol) = 17.03\ g/mol$
Calculate the moles of $NH_3$ produced: $Moles\ of\ NH_3 = \frac{mass\ of\ NH_3}{molar\ mass\ of\ NH_3} = \frac{11.27\ g}{17.03\ g/mol} = 0.662\ mol$
Use the stoichiometric coefficients to find the moles of $H_2$ needed: $Moles\ of\ H_2 = 3 \times Moles\ of\ NH_3 = 3 \times 0.662\ mol = 1.986\ mol$
Calculate the grams of $H_2$ needed: $Mass\ of\ H_2 = Moles\ of\ H_2 \times Molar\ mass\ of\ H_2 = 1.986\ mol \times 2.02\ g/mol = 4.01\ g$

Therefore, 4.01 grams of $H_2$ are needed to produce 11.27 grams of $NH_3$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.