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(c) (i) Ammonium nitrate is another salt of ammonia that can be used as a fertiliser. It can be made from the reaction of ammonia with nitric acid. The balanced equation for the reaction is NH_(3)+HNO_(3)arrow NH_(4)NO_(3) A manufacturer reacts 500g of ammonia with 1000g of nitric acid. Determine the limiting reagent in the reaction. (relative formula masses NH_(3)=17.0,HNO_(3)=63.0) Show your working limiting reagent= __ (ii) A student wants to carry out the same reaction in the laboratory on a smaller scale. The student uses 50cm^3 of a 0.3moldm^-3 solution of ammonia. Calculate the mass of ammonia in the solution in grams. (relative formula mass NH_(3)=17.0) Show your working

Question

(c) (i) Ammonium nitrate is another salt of ammonia that can be used as a fertiliser.
It can be made from the reaction of ammonia with nitric acid.
The balanced equation for the reaction is
NH_(3)+HNO_(3)arrow NH_(4)NO_(3)
A manufacturer reacts 500g of ammonia with 1000g of nitric acid.
Determine the limiting reagent in the reaction.
(relative formula masses NH_(3)=17.0,HNO_(3)=63.0)
Show your working
limiting reagent= __
(ii) A student wants to carry out the same reaction in the laboratory on a
smaller scale.
The student uses 50cm^3 of a 0.3moldm^-3 solution of ammonia.
Calculate the mass of ammonia in the solution in grams.
(relative formula mass NH_(3)=17.0)
Show your working

(c) (i) Ammonium nitrate is another salt of ammonia that can be used as a fertiliser. It can be made from the reaction of ammonia with nitric acid. The balanced equation for the reaction is NH_(3)+HNO_(3)arrow NH_(4)NO_(3) A manufacturer reacts 500g of ammonia with 1000g of nitric acid. Determine the limiting reagent in the reaction. (relative formula masses NH_(3)=17.0,HNO_(3)=63.0) Show your working limiting reagent= __ (ii) A student wants to carry out the same reaction in the laboratory on a smaller scale. The student uses 50cm^3 of a 0.3moldm^-3 solution of ammonia. Calculate the mass of ammonia in the solution in grams. (relative formula mass NH_(3)=17.0) Show your working

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GrahamAdvanced · Tutor for 1 years

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(c)(i) Based on our calculations from Step 1 \(NH_3\) forms approximately 29.4 mol and \(HNO_3\) forms approximately 15.9 mol, So, based on the ratio established in the balance equation,ottfried Paul\(NH_3\) will be consumed before \(HNO_3\). Hence, the limiting reagent = \(NH_3\)<br /><br />(ii) From the calculations performed during Step 3: for 0.05*0.3 = 0.015 mol. Therefore, the mass of the ammonia in 50 cm^3 of 0.03 M solution is given by: 0.015 mol * 17.0 gm/mol (molar mass of ammonia) = 0.255 grams.

Explain

## Step 1:<br />First, let's find out the moles of the given masses of ammonia (\(NH_3\)) and nitric acid (\(HNO_3\)).<br /> <br />We know that,<br /> <br />### **Number of moles = \(\frac{{given mass}}{{molar mass}}\)**<br /> <br />So, for \(NH_3\):<br /> Number of moles = \(\frac{{500}}{{17.0}}\)<br /> <br />And for \(HNO_3\): <br />Number of moles = \(\frac{{1000}}{{63.0}}\)<br /><br />## Step 2:<br />The molar ratio between \(NH_3\) and \(HNO_3\) in the balanced equation is 1:1. From the above step, if the amount of \(NH_3\) moles is lesser than the \(HNO_3\) moles, then \(NH_3\) becomes the limiting factor i.e., \(NH_3\) is the limiting reagent.<br /><br />## Step 3:<br />Next, for the student’s laboratory practical, let’s find mass of ammonia in 0.3 mol dm-3 solution that is housed in a 50 cm^3 container.<br /><br />This problem can be solved using large and small scales. On a large scale, 1 dm^3 = 0.3 mol. On a small scale, 0.05 dm^3 (since 1 dm^3 = 1000 cm^3, 50 cm^3 = 0.05 dm^3) will yield (0.05*0.3) moles of \(NH_3\). <br />Now using this information, calculate the mass using the number of moles, multiplied by the molar mass of \(NH_3\):<br /><br />### **Mass = Number of moles * Molar mass.**
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