C(5) + O2(g)
Hope this helped
Answer:
MnO₄⁻ + 2 H₂O + 3 e⁻ ⇒ MnO₂ + 4 OH⁻
2 moles of H₂O are on the reactant side.
Explanation:
Let´s consider the following half-reaction.
MnO₄⁻ ⇒ MnO₂
This reduction of permanganate to manganese dioxide happens in a neutral medium. The oxidation number of Mn changes from +7 to +4. When we are in a neutral medium, we can balance the reaction using the ion-electron method either as we were in acid or in basic medium. In this case, to satisfy the condition is the task (<em>water on the reactant side</em>) we will balance this half-reaction as if it were in basic media. Since Mn is already balanced, we have to balance O. We will add as many moles of water as excess of oxygen there is on 1 side, and the double of OH⁻ on the other side. Finally, we add as many electrons as we need to balance it electrically.
MnO₄⁻ + 2 H₂O + 3 e⁻ ⇒ MnO₂ + 4 OH⁻
The number Beryllium chloride consumed is calculated as follows
write equation for reaction
that is BeCl2 + Mg -----> MgCl2 +Be
find the moles of MgCl2 produced
moles =mass/molar mass of MgCl2
= 987.5g/95 = 10.39 moles
by use of mole ratio between BeCl2 to Mgcl2 which 1:1 therefore the moles of BeCl2 is also 10.39moles
mass of BeCl2 = moles x molar mass
that is 10.39 x80 = 831.2
