If 0.1800 g of impure soda ash (Na₂CO₃) is titrated with 15.66 mL of 0.1082 M HCl, what is the percent purity of the soda ash?
Na₂CO₃(aq) + 2 HCl(aq) ? 2 NaCl(aq) + H₂O( $\ell$ ) + CO₂(g)

Question

If 0.1800 g of impure soda ash (Na₂CO₃) is titrated with 15.66 mL of 0.1082 M HCl, what is the percent purity of the soda ash? Na₂CO₃(aq) + 2 HCl(aq) ? 2 NaCl(aq) + H₂O( $\ell$ ) + CO₂(g)

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The Answer of If 0.1800 g of impure soda ash (Na₂CO₃) is titrated with 15.66 mL of 0.1082 M HCl, what is the percent purity of the soda ash? Na₂CO₃(aq) + 2 HCl(aq) ? 2 NaCl(aq) + H₂O( $\ell$ ) + CO₂(g)

If 01800 G of Impure Soda Ash (Na2CO3) Is Titrated with with 15.66

If 01800 G of Impure Soda Ash (Na₂CO₃) Is Titrated with with 15.66