How many grams of H_2 are needed to react with 28 grams of O_2, given the reacti

Question

How many grams of H_2 are needed to react with 28 grams of O_2, given the reaction 2 H_2 (g) + O_2 (g) rightarrow 2 H_2 O(l) How many mL of 0.25 M CaBr_2 are needed to react with 25 ml of 0.4 M U_3 PO_4. given the balanced equation 2 U_3 PO_4 (aq) + 3 CaBr_2 (aq) rightarrow 6 LiBr_2 (aq) + Ca_3 (PO_4)_2 (s) How many grams of Ca_3 (PO_4)_2 precipitate can form by reacting 258.6 mL of 1.8 M CaBr_2 with an excess amount of Li_3 PO_4, given the balanced equation. 2 Li_3 PO_4 (aq) + 3 CaBr_2 (aq) rightarrow 6 LiBr_2 (aq) + Ca_3 (PO_4)_2 (s)

Explanation / Answer

Q1. from balanced reaction;Two mole of H2 reacting with 1 mole of O2.

No.f mole of oxygen(o2) in 28g = 28/32

= 0.875 mol

0.875 mole of O2 need H2 mole = 0.875 x 2= 1.75mol

mass of H2 = No.of mole x molar mass of H2

= 1.75 x 2

= 3.5g

Q2. Given number of mole of Li3PO4 = Molarity(M) x Volume in liter

=0.4 x 0.025

= 0.01 mol

from balanced reaction 2mole of Li3PO4 need 3 mole of CaBr

1 mole of Li3PO4 need mole of CaBr to react = 3/2 mole

0.01 mole of Li3PO4 need mole of CaBr2 to react = 3/2 x 0.01 mol

= 0.015 mol

Volume of CaBr2 = number of mole / Molarity of CaBr2

= 0.015 / 0.25

= 0.06 L

= 60 ml

Q3.

number of mole of CaBr2 = 1.8 x 0.2586

= 0.4655 mol

from balanced reaction 3mole of CaBr2 form 1 mole of precipitate.Ca3(PO4)2

1 mole of  CaBr2 form mole of precipitate.Ca3(PO4)2 = 1/3 mole

0.4655 mole of CaBr2 forms mole of precipitate.Ca3(PO4)2 = 1/3 x 0.4655 mole

= 0.1552 mol

grams of precipitate.Ca3(PO4)2 = number of mole x molar mass

= 0.1552 x 310.177

=48.139 grams

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