in a reaction 20g aqueous solution of compound A was mixed with 20g aqueous solution of compound of B. if 9.8g of solid product C was formed then the mass of the other product D will be (Assume that data is according to the law of conservation of mass)
Answers
Explanation:
Sodium bicarbonate, #NaHCO_3#, and acetic acid, #CH_3COOH#, will react to produce sodium acetate, #CH_3COONa#, and carbonic acid, #H_2CO_3#.
The carbonic acid will then decompose to give water and carbon dioxide, #CO_2#.
#NaHCO_(3(s)) + CH_3COOH_((aq)) -> CH_3COONa_((aq)) + underbrace(H_2O_((l)) + CO_(2(g)))_(color(blue)(H_2CO_3)#
So, you've got 8.4 g of sodium bicarbonate reacting with 20g of acetic acid, and producing 4.4 g of carbon dioxide. The first thing you need is work backwards from the mass of #CO_2# produced to determine what is the mass of acetic acid that reacted.
To do this, use the compounds' molar masses to calculate how many moles of each you have
#4.4cancel("g") * ("1 mole "CO_2)/(44.01cancel("g")) ~= "0.10 moles "# #CO_2#
#8.4cancel("g") * ("1 mole "NaHCO_3)/(84.007cancel("g")) ~= "0.10 moles "# #NaHCO_3#
According to the #1:1# mole ratio that exists between sodium bicarbonate, acetic acid, and carbon dioxide, the number of moles of acetic acid that reacted is
#0.10cancel("moles "CO_2) * ("1 mole "CH_3COOH)/(1cancel("mole "CO_2)) = "0.10 moles "# #CH_3COOH#
Therefore, the mass that reacted was
#0.10cancel("moles "CH_3COOH) * "60.005 g"/(1cancel("mole " CH_3COOH)) ~= "6.0 g "# #CH_3COOH#
Therefore, the total mass of the reactants wil be
#m_"reactants" = 8.4 + 6.0 = "14.4 g"#
According to the law of mass conservation, the total mass of the products must be equal to the total mass of the reactants., which means that
#m_"products" = m_"reactants" = "14.4 g"#
Therefore,
#m_"products" = m_(CO_2) + m_"residue"#, or
#m_"residue" = m_"products" - m_(CO_2) = 14.4 - 4.4 = "10.0 g"#
#overbrace(NaHCO_(3(s)))^(color(blue)("8.4 g")) + overbrace(CH_3COOH_((aq)))^(color(blue)("6.0 g")) -> overbrace(CH_3COONa_((aq)) + H_2O_((l)))^(color(green)("10.0 g")) + overbrace(CO_(2(g)))^(color(green)("4.4 g"))#