Hard water contains $MgSO X_{4}$ . Which option gives a correctly balanced equation for scum formation with sodium stearate and also the correct amount of scum obtained from $0.30 mol$ of sodium stearate (excess $Mg X^{2 +}$ )?

A $2 C X_{17} H X_{35} COONa + MgSO X_{4} (C X_{17} H X_{35} COO) X_{2} Mg + Na X_{2} SO X_{4}$ ; $0.15 mol$
B $2 C X_{17} H X_{35} COONa + MgSO X_{4} (C X_{17} H X_{35} COO) X_{2} Mg + Na X_{2} SO X_{4}$ ; $0.30 mol$
C $C X_{17} H X_{35} COONa + MgSO X_{4} (C X_{17} H X_{35} COO) X_{2} Mg + Na X_{2} SO X_{4}$ ; $0.30 mol$
D $2 C X_{17} H X_{35} COONa + MgSO X_{4} (C X_{17} H X_{35} COO) X_{2} Mg + Na X_{2} SO X_{4}$ ; $0.60 mol$

Answer: A

Each magnesium stearate needs two stearate ions, so the balanced equation has 2 C17H35COONa. The ratio is 2 mol soap to 1 mol scum, so 0.30 mol soap gives 0.30/2 = 0.15 mol. The other options have an unbalanced equation or a wrong ratio.

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Hard water contains MgSOX4​. Which option gives a correctly balanced equation for scum formation with sodium stearate and also the correct amount of scum obtained from 0.30 mol of sodium stearate (excess MgX2+)?

Answer: A

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Hard water contains $MgSO X_{4}$ . Which option gives a correctly balanced equation for scum formation with sodium stearate and also the correct amount of scum obtained from $0.30 mol$ of sodium stearate (excess $Mg X^{2 +}$ )?