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Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H3PO4 react according to the following chemical equation.\(2Cr + 2{H_3}P{O_4} \to 2CrP{O_4} + 3{H_2}\). Determine the limiting reactant.

Short Answer

Expert verified

Cr is the limiting reactant.

Step by step solution

01

Determine the number of moles

Find the number of moles of \({H_3}P{O_4}\).

Mol\({H_3}P{O_4} = 0.50\,mol\,Cr\,\left( {\frac{{2\,mol\,{H_3}P{O_4}}}{{2\,mol\,Cr}}} \right) = 0.50\,mol\,{H_3}P{O_4}\).

02

Determine the limiting reactant

Only 0.50 mol of\({H_3}P{O_4}\)is required but we have 0.75 mol of\({H_3}P{O_4}\)so, \({H_3}P{O_4}\)is in excess amount. Thus, Cr is the limiting reactant.

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Most popular questions from this chapter

What volume of 0.08892 M HNO3 is required to react completely with 0.2352 g of potassium hydrogen phosphate?

\(2{\rm{HN}}{{\rm{O}}_{3({\rm{aq}})}}{\rm{ + }}{{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ }} \to {\rm{ }}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_{4({\rm{aq)}}}}{\rm{ + 2KN}}{{\rm{O}}_{3({\rm{aq}})}}\)

Diatomic chlorine and sodium hydroxide(lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen via the electrolysis of brine according to the following unbalanced equation:

\[NaCl\left( {aq} \right) + {H_2}O\left( l \right) \to NaOH\left( {aq} \right) + {H_2}\left( g \right) + C{l_2}\left( g \right).\]

Write balanced molecular, complete ionic, and net ionic equations for this process.

Complete and balance the following acid-base equations:

(a) A solution ofHClO4 is added to a solution of LiOH

(b) AqueousH2SO4 reacts withNaOH

(c)Ba(OH)2reacts withHF gas.

How many molecules of C2H4Cl2 can be prepared from 15 C2H4molecules and 8 Cl2molecules?

Balance the following equations:

\(\begin{array}{l}\;(a)\;PC{l_5}\left( s \right) + {H_2}O\left( l \right) \to POC{l_3}\left( l \right) + HCl\left( {aq} \right)\\\left( b \right)\,Cu\left( s \right) + HN{O_3}\left( {aq} \right) \to Cu{\left( {N{O_3}} \right)_2}\left( {aq} \right) + {H_2}O\left( l \right) + NO\left( g \right)\\\left( c \right){H_2}\left( g \right) + {I_2}\left( g \right) \to HI\left( s \right)\\\left( d \right)\,Fe\left( s \right) + {O_2}\left( g \right) \to F{e_2}{O_3}\left( s \right)\\\left( e \right)\,Na\left( s \right) + {H_2}O\left( l \right) \to NaOH\left( {aq} \right) + {H_2}\left( g \right)\\\left( f \right)\,{\left( {N{H_4}} \right)_2}C{r_2}{O_7}\left( s \right) \to C{r_2}{O_3}\left( s \right) + {N_2}\left( g \right) + {H_2}O\left( g \right)\\\left( g \right){P_4}\left( s \right) + C{l_2}\left( g \right) \to PC{l_3}\left( l \right)\\\left( h \right)PtC{l_4}\left( s \right) \to Pt\left( s \right) + C{l_2}\left( g \right)\end{array}\)

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