If 0.5 mole of BaCl2 is mixed with0.2 mole of Na3PO4 the maximum number of mole Ba3(PO4)2 that can be formed is?

Md Ariful Islam

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Reaction: 3BaCl2 + 2Na3PO4 = Ba3(PO4)2 + 6NaCl The mole ratio between BaCl2 and Na3PO4 is n(BaCl2):n(Na3PO4)=3:2 The ratio of these compounds in the given mixture is 0.5/0.2 = 5:2 gt; 3:2. Hence the BaCl2 is in excess and the Na3PO4 is the limiting reactant. The 1 mole of Ba3(PO4)2 occurs for each two moles of Na3PO4, hence the amount of Ba3(PO4)2 n[Ba3(PO4)2] = 1/2 * 0.2 = 0.1 mol.
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If 16 moles of Ba3(PO4)2 are produced, how many moles of Ba(OH)2 are required for the reaction

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Consider the following balanced equation: 3NH4NO3 + Na3PO4 → (NH4)3PO4 + 3NaNO3 30 g NH4NO3 and 50 g Na3PO4 are allowed to react -What is the maximum amount of each product that can be prepared?

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H3PO4+Ba(OH)2->Ba3+(PO4)2+H2O The big #s are saposed to be sub #s and its a Limited reactant problem

2H3PO4 + 3Ba(OH)2 --gt; Ba3(PO4)2 + 6H2O

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Ca5(PO4)3F + 5H2SO4 = 3H3PO4 + 5CaSO4 + HF if 24.5 g Ca(PO4)3F is mixed with 89.7 g H2SO4, how much H3PO4 will form? NEED IT TODAY

The quantity of Ca5(PO4)3F:(24.5 g) / (504.3 g/mol) = 0.0486 molThe quantity of H2SO4:(89.7 g) / (98.1 g/mol) = 0.9144 molThe ratio of these reactants must be 1:5, so H2SO4...

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1)since ba3(po4)2 is a very finely divided solid some is lost during filtration this is specially true if a coarse filter paper is used how would this affect the percent error of the experiment?

Solution: Coarse filter paper can’t be used to filter ultrafine precipitation. This can lead to serious losses of the product and ultimately to an increase in the percentage error of...

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When 3.802 g of Na3 PO4 ·12H2O react with excess BaCl2 ·2H2O, how many moles of Ba3 (PO4 ) 2 would be produced?

2Na3PO4·12H2O + 3BaCl2·2H2O => Ba3(PO4)2 + 6NaCl + 30H2OM(Na3PO4·12H2O) = 380.12 g/mol;n=m/M;n(Na3PO4·12H2O) = (3.802 g)/(380.12 g/mol) = 0.01 moles;Proportion according to the reaction:2 moles (Na3PO4·12H2O) — 1 mole (Ba3(PO4)2)0.01 moles...

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According to the equation, moles AgNO3 = moles AgCl.So the theoretical yield of this reaction can be calculated as: mtheor(AgCl) = (m(AgNO3) / M(AgNO3)) x M (AgCl) = (3.38g /...

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How many grams of BaCl2 should be added to 300 g H2O so that the freezing point of the solution is lowered to -8.3°C? Assume that the BaCl2 completely dissociates in the solution.

92.92 grams

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