Question

question 1
metal carbonates react with acid to produce a salt, water and carbon
dioxide gas as seen in the reaction below.
mathrm{na}_{2} mathrm{co}_{3}(s)+2 mathrm{hcl}(a q)
ightarrow 2 mathrm{nacl}(a q)+mathrm{h}_{2} mathrm{o}(prime)+mathrm{co}_{2}(g)
if ( 472 mathrm{~ml} ) of ( 2.24 mathrm{m} mathrm{hcl} ) reacts with an excess amount of ( mathrm{na}_{2} mathrm{co}_{3} ), what volume of ( mathrm{co}_{2} ) gas at stp
will be produced?

question 2
the reaction between hydrogen gas and nitrogen gas produces ammonia
gas according to the reaction:
mathrm{n}_{2}(g)+3 mathrm{h}_{2}(g)
ightarrow 2 mathrm{nh}_{3}(g)
what volume of hydrogen gas is needed to react with ( 20.8 mathrm{~l} ) of nitrogen
gas if both gases are at stp?
if ( 21.2 mathrm{~l} ) of hydrogen gas reacts with excess nitrogen gas at stp, what volume of ammonia gas
will be formed?

Explanation:

Question 1

Step1: Calculate moles of HCl

Use the formula \(n = M\times V\) (where \(n\) is moles, \(M\) is molarity, \(V\) is volume in liters). Given \(M = 2.24\space M\) and \(V=472\space mL=0.472\space L\), then \(n_{HCl}=2.24\times0.472 = 1.05728\space mol\)

Step2: Find moles of \(CO_2\)

From the balanced equation \(Na_2CO_3(s)+2HCl(aq)\to2NaCl(aq)+H_2O(l)+CO_2(g)\), the mole ratio of \(HCl\) to \(CO_2\) is \(2:1\). So \(n_{CO_2}=\frac{n_{HCl}}{2}=\frac{1.05728}{2}=0.52864\space mol\)

Step3: Calculate volume of \(CO_2\) at STP

At STP (\(T = 273\space K\), \(P = 1\space atm\)), \(1\space mol\) of an ideal gas has a volume \(V_m = 22.4\space L/mol\). Using \(V=n\times V_m\), \(V_{CO_2}=0.52864\times22.4 = 11.84\space L\)

Step1: Use mole - volume ratio

From the balanced equation \(N_2(g)+3H_2(g)\to2NH_3(g)\), the volume ratio of \(N_2\) to \(H_2\) (since at STP, mole ratio = volume ratio for gases) is \(1:3\). Given \(V_{N_2}=20.8\space L\), then \(V_{H_2}=3\times V_{N_2}\)

Step2: Calculate volume of \(H_2\)

\(V_{H_2}=3\times20.8 = 62.4\space L\)

Step1: Use mole - volume ratio

From the balanced equation \(N_2(g)+3H_2(g)\to2NH_3(g)\), the volume ratio of \(H_2\) to \(NH_3\) is \(3:2\). Given \(V_{H_2}=21.2\space L\)

Step2: Calculate volume of \(NH_3\)

\(V_{NH_3}=\frac{2}{3}\times V_{H_2}\), \(V_{NH_3}=\frac{2}{3}\times21.2=\frac{42.4}{3}=14.13\space L\)

Answer:

\(11.84\)

Question 2

Part 1