 # Write the balanced chemical equation for each reaction. a) Solid c Francisca Rodden 2021-12-17 Answered
Write the balanced chemical equation for each reaction.
a) Solid copper reacts with solid sulfur to form solid copper(I) sulfide.
b) Solid iron (III) oxide reacts with hydro gen gas to form solid iron and liquid water.
c) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas.
d) Gaseous ammonia (NH3) reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water.
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Step 1
First equation is unbalanced because on the right side of the equation is 2 atoms of copper, and on the left is one. Second one is balanced.
$Cu\left(s\right)+S\left(s\right)=C{u}_{2}S\left(s\right)$
$2Cu\left(s\right)+s\left(s\right)=C{u}_{2}S\left(s\right)$
Step 2
Second one is balanced equation. There is 2 atoms of Fe, 6 of H and 3 of O.
$F{e}_{2}{O}_{3}\left(s\right)+{H}_{2}\left(g\right)=Fe\left(s\right)+{H}_{2}O\left(l\right)$
$F{e}_{2}{O}_{3}\left(s\right)+3{H}_{2}\left(g\right)=2Fe\left(s\right)+3{H}_{2}O\left(l\right)$
Step 3
Second one is balanced equation. There is 2 atoms of S and 6 atoms of O.
$S{O}_{2}\left(g\right)+{O}_{2}\left(g\right)=S{O}_{3}\left(g\right)$
$2S{O}_{2}\left(g\right)+{O}_{2}\left(g\right)=2S{O}_{3}\left(g\right)$
Step 4
There is 4 atoms of N, 10 of O and 12 of H in balanced equation.
$N{H}_{3}\left(g\right)+{O}_{2}\left(g\right)=NO\left(g\right)+{H}_{2}O\left(g\right)$
$N{H}_{3}\left(g\right)+5{O}_{2}\left(g\right)=4NO\left(g\right)+6{H}_{2}O\left(g\right)$
###### Not exactly what you’re looking for? Annie Levasseur
Part A
Write the unbalanced equation with the correct formulas and states.
Balance Cu with a coefficient of 2 for Cu. All elements are now balanced.
$Cu\left(s\right)+S\left(s\right)⇒C{u}_{2}S\left(s\right)$
$2Cu\left(s\right)+S\left(s\right)⇒C{u}_{2}S\left(s\right)$
Part B
Write the unbalanced equation with th ecorrect formulas and states.
Use the following steps to balance the equation:
1) Balance O with a coefficient of 3 for ${H}_{2}O$
2) Balance H with a coefficient of 3 for ${H}_{2}$
3) Balance Fe with a coefficient of 2 for Fe
$F{e}_{2}{O}_{3}\left(s\right)+{H}_{2}\left(g\right)⇒Fe\left(s\right)+{H}_{2}O\left(l\right)$
$F{e}_{2}{O}_{3}\left(s\right)+3{H}_{2}\left(g\right)⇒2Fe\left(s\right)+3{H}_{2}O\left(l\right)$
Part C
Write the unbalanced equation with the correct formulas and states
Use the following steps to balance the equation:
1) Balance O with a coefficient of $\frac{1}{2}$ for ${O}_{2}$. All elements are now balanced.
2) Multiply both sides of the equation by 2 in order to remove fractional coefficients.
$S{O}_{2}\left(g\right)+{O}_{2}\left(g\right)⇒S{O}_{3}\left(g\right)$
$S{O}_{2}\left(g\right)+\frac{1}{2}{O}_{2}\left(g\right)⇒S{O}_{3}\left(g\right)$
$2S{O}_{2}\left(g\right)+{O}_{2}\left(g\right)⇒2S{O}_{3}\left(g\right)$
Part D:
Write the unbalanced equation with the correct formulas and states.
Use the following steps to balance the equation:
1) Balance H with coefficients of 2 for $N{H}_{3}$ and 3 for ${H}_{2}O$
2) Balance N with a coefficient of 2 for NO
3) Balance O with a coefficient of $\frac{5}{2}$ for ${O}_{2}$
4) Nultiply both sides of the equation by 2 in order to remove fractional coefficients.
$N{H}_{3}\left(g\right)+{O}_{2}\left(g\right)⇒NO\left(g\right)+{H}_{2}O\left(g\right)$
$2N{H}_{3}\left(g\right)+\frac{5}{2}{O}_{2}\left(g\right)⇒2NO\left(g\right)+3{H}_{2}O\left(l\right)$
$4N{H}_{3}\left(g\right)+5{O}_{2}\left(g\right)⇒4NO\left(g\right)+6{H}_{2}O\left(g\right)$