Question

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O. Suppose 24.1 g of hydrochloric acid is mixed with 51. g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

$0.9[g]$

Explanation:

The reaction can be represented with a following equation:

$HCI+NaOH=NaCl+H$₂$O$

According  to  the  equation,  equal  amounts  of  chemical  agents  and  resultant  products  (in moles) participate in this reaction.The amount of hydrochloric acid is as follows:

$V_{HCI} =\frac{M_{HCI} }{M_{HCI} } =\frac{M_{HCI} }{A_{H}+A_{CI} } =\frac{1.8}{1+35.46} =0.05[Moles],$

where m is the substance mass, A is the atomic mass of an element.The amount of sodium hydroxide is as follows:

$V_{NaOH} =\frac{m_{NaOH} }{M_{NaOH} } =\frac{m_{NaOH} }{A_{Na} +A_{o}+A_{H} }=\frac{3.66}{23+16+1} =0.09[moles],$

Since  the  amount  of  sodium  hydroxide  exceeds  the  amount  of  hydrochloric  acid,  the maximum amount of each substance participating in the reaction is $V_{HCI} =0.05(moles)$

Therefore, the maximum amount of water that can be produced is as follows:

$m_{H2O} =M_{H2O} =(2A_{H}+A_{O} )*v=(2*1+16)*0.05=0.9[g]$

Answer: $0.9[g]$