Use the balanced chemical equation below to answer the questions that follow: Question 3
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Marked out of
2Na, PO, +3MgCl2 →
6NaCl + Mg,(PO)2
*Round all answers to 2 decimal places

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2