Question:
Calculate the maximum number of mole of Ba3(PO4)2 when 0.6 mole of BaCl2 is mixed with 0.6 mole of Na3PO4.
Solution:
According to the stoichiometric ratio, sodium phosphate is in excess; hence, barium chloride is the limiting reactant.
n(Ba3(PO4)2) (mole) = n(BaCl2)*2/3 = (0.6/3)*2 = 0.4 mol 

Answer: 0.4 mol

Get your answer at https://www.chemtutor.me/
See more solved problems at https://www.chemtutor.me/problems/
Also check out our free chemistry tools at https://www.chemtutor.me/tools/

Comments

Post a Comment

Popular posts from this blog

Question: (a) The reaction between hydrogen, H 2 (g), and nitrogen monoxide, NO(g), has the following rate equation, rate = k[H 2 (g)][NO(g)] 2 , The overall equation for the reaction is, 2H 2 (g) + 2NO(g) → N 2 (g) + 2H 2 O(l), and the reaction takes place by a two step mechanism. (i)     What is the meaning of reaction mechanism? (ii)   Explain what a rate determining step is. (iii)              Giving your reasoning, write two equations that could describe the two mechanistic steps for the above reaction. It is important to identify the rate determining step. Answers: (i)                  A reaction mechanism is the step by step sequence of elementary reactions through which the overall chemical reaction occurs. (ii)                A rate determining step is the slowest step of complex reaction, and this step determines the overall reaction rate. (iii)              Based on the rate equation we can conclude that rate determining step is    H 2 (g) + 2NO(g)
Read more
Question   16. 2.4 kg of carbon is made to react with 1.35 kg of aluminium to form Al4C3. Calculate the maximun amount (in kg) of aluminium carbide formed Answer The chemical equation is: 4Al + 3C = Al 4 C 3 Amount of substance, n , for carbon is:   n = m/M(C) n = 2400/12 = 200 mole Amount of substance for aluminium is:   n = 1350/27 = 50 mole Therefore Al is the limiting reagent.   Amount of substance for Al 4 C 3 will be:   (1/4) n(Al) = 50/4 = 12.5 mole And the weight of Al 4 C 3 will be: m = n*M   m =   12.5*144 = 1800 g (1.8 kg) Answer : 1.8 kg Get your answer at  https://www.chemtutor.me/ See more solved problems at  https://www.chemtutor.me/problems/ Also check out our free chemistry tools at  https://www.chemtutor.me/tools/
Read more
Question: 600 ml of ozonised oxygen at STP was found to weight one gram. What is the volume of O3 in the ozonoised oxygen ? (1)                200 ml (2)                150 ml (3)                100 ml (4)                50 ml Answer: From the ideal gas law we can find the expression for the molar mass: pV = nRT n = m/M pV = mRT/M M = mRT/pV = 1∙8.314∙273/(100000∙0.0006) = 37.8 g/mol M(O 3 ) = 48 M(O 2 ) = 32 The average molar mass of gas can be expressed as the sum of molar masses multiplied by mole fraction. Let x denote the mole fraction of ozone, so the mole fraction oxygen is 1-x. Now we can write the expression for x: 48x + 32(1-x) = 37.8 x = 0.3625 The volume of ozone is the total volume multiplied by the ozone mole fraction: V(O 3 ) = x∙V = 217.5 ml ≈ 200 ml So, the correct option is (1). Get your answer at  https://www.chemtutor.me/ See more solved problems at  https://www.chemtutor.me/problems/ Also check out our free chemistry tools at  https:/
Read more