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August 23, 2014

August 23, 2014

Posted by **Kelly** on Tuesday, October 5, 2010 at 12:29am.

For each of the reactions shown, calculate the mass (in grams) of the product formed when 10.7 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

2K(s)+Cl_2(g) ->2K Cl(s)

2K(s)+ Br_2(l) -> 2KBr(s)

- Chemistry Help -
**DrBob222**, Tuesday, October 5, 2010 at 12:43ammoles Cl2 = 10.7/about 71 = about 0.15

Convert moles Cl2 to moles KCl.

0.15 moles Cl2 x (2moles KCl/1 mole Cl2) = 0.15 x (2/1) = 0.30 moles.

g KCl = moles x molar mass. = about 0.30 x 74.55 = about 22.5 grams KCl.

You must go through and calculate more precisely than I. Be careful with the number of significant figures.

I obtained an exact value of

22.501 which will round to 22.5 grams to three s.f.

The KBr problem is done the same way. My number for that problem is

15.935 grams which rounds to 15.9 g to three s.f.

- Chemistry Help -
**Kelly**, Tuesday, October 5, 2010 at 1:08amThank you so much for your help, for the following question i am getting wrong answers answell. for the first one i got 42.7 and the 2nd I got 19.8. Could you help? Thank you for help in advance

4Cr(s) = 3O_2(g) -> 2Cr_2O_3(s)

2Cr(s)+O_2(g)->2SrO(s)

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